Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 1 :

$If \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + constant,$ $then the maximum value of a \sin x + b \cos x is:$ [2024]

$\sqrt{41}$
$\sqrt{42}$
$\sqrt{40}$
$\sqrt{39}$

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(3)

We have, $\int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + c$

Let $I = \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \int \frac{\sec^{2} x}{a^{2} \tan^{2} x + b^{2}} d x$

Let $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$∴$ $I = \int \frac{d t}{{(a t)}^{2} + b^{2}}$

$= \frac{1}{b a} \tan^{- 1} (\frac{a t}{b}) + c = \frac{1}{b a} \tan^{- 1} (\frac{a}{b} \tan x) + c$

$\Rightarrow a b = 12 and \frac{a}{b} = 3$

$\Rightarrow a^{2} = 36 and b^{2} = 4$

Now, $- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$

Required maximum value $= \sqrt{36 + 4} = \sqrt{40}$

Q 2 :

Let $I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x .$ If $I (0) = 3,$ then $I (\frac{π}{12})$ is equal to [2024]

$6 \sqrt{3}$
$3 \sqrt{3}$
$\sqrt{3}$
$2 \sqrt{3}$

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(2)

$I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x = \int \frac{6 c o s e c^{2} x}{{(1 - \cot x)}^{2}} d x$

Put $1 - \cot x = t \Rightarrow c o s e c^{2} x d x = d t$

$\Rightarrow I = \int \frac{6 d t}{t^{2}} = \frac{- 6}{t} + C \Rightarrow I = \frac{- 6}{1 - \cot x} + C$

Now, $I (0) = 3 \Rightarrow 3 = C$

$\Rightarrow I (x) = 3 - \frac{6}{1 - \cot x} \Rightarrow I (\frac{π}{12}) = 3 - \frac{6}{1 - \cot \frac{π}{12}}$

$= 3 - \frac{6}{1 - (2 + \sqrt{3})} = \frac{- 3 - 3 \sqrt{3} - 6}{- 1 - \sqrt{3}} = \frac{- 9 - 3 \sqrt{3}}{- 1 - \sqrt{3}}$

$= \frac{(9 + 3 \sqrt{3}) (1 - \sqrt{3})}{- 2} = \frac{9 - 9 \sqrt{3} + 3 \sqrt{3} - 9}{- 2} = 3 \sqrt{3}$

Q 3 :

Let $\int \frac{2 - \tan x}{3 + \tan x} d x = \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C,$ where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to: [2024]

7
3
1
4

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(4)

Let $I = \int \frac{2 - \tan x}{3 + \tan x} d x = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} d x$

$Write Numerator = λ (Differentiation of Denominator) + μ (Denominator)$

$2 \cos x - \sin x = λ (- 3 \sin x + \cos x) + μ (3 \cos x + \sin x)$

$\Rightarrow λ + 3 μ = 2 \dots (i)$

$and - 3 λ + μ = - 1 \dots (ii)$

$From equations (i) and (ii), λ = μ = \frac{1}{2}$

$I = \frac{1}{2} \int \frac{- 3 \sin x + \cos x}{3 \cos x + \sin x} d x + \frac{1}{2} \int \frac{3 \cos x + \sin x}{3 \cos x + \sin x} d x$

$= \frac{1}{2} \int \frac{d t}{t} + \frac{1}{2} \int d x [Putting 3 \cos x + \sin x = t in first integral \Rightarrow (- 3 \sin x + \cos x) d x = d t]$

$= \frac{1}{2} \ln (t) + \frac{1}{2} x + C$

$= \frac{1}{2} \ln$ $| 3 \cos x + \sin x |$ $+ \frac{1}{2} x + C$

$= \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C (∵Given)$

$\Rightarrow α = 1, β = 1, γ = 3$

$Now, α + \frac{γ}{β} = 1 + \frac{3}{1} = 4$

Q 4 :

The integral $\int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$ is equal to: [2024]

$\log_{e}$ $(| \tan^{- 1} (x^{3} + \frac{1}{x^{3}}) |) + C$
$\log_{e}$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{3} + C$
$\log_{e}$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{\frac{1}{3}} + C$
$\log_{e}$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{\frac{1}{2}} + C$

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(3)

Let $I = \int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$

Put $\tan^{- 1} (x^{3} + \frac{1}{x^{3}}) = u$

$\Rightarrow \frac{1}{1 + {(x^{3} + \frac{1}{x^{3}})}^{2}} \cdot 3 (x^{2} - \frac{1}{x^{4}}) d x = d u$

$\Rightarrow (\frac{x^{6}}{x^{12} + 3 x^{6} + 1} \times \frac{x^{6} - 1}{x^{4}}) d x = \frac{d u}{3}$

$\Rightarrow I = \frac{1}{3} \int \frac{d u}{u} = \frac{\ln | u |}{3} = \ln$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{\frac{1}{3}} + C$

Q 5 :

$For x \in (- \frac{π}{2}, \frac{π}{2}), if y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x,$ $and \lim_{x \to {(\frac{π}{2})}^{-}} y (x) = 0, then y (\frac{π}{4}) is equal to$ [2024]

$- \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}})$
$\tan^{- 1} (\frac{1}{\sqrt{2}})$
$\frac{1}{2} \tan^{- 1} (\frac{1}{\sqrt{2}})$
$\frac{1}{\sqrt{2}} \tan^{- 1} (- \frac{1}{2})$

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(4)

Let $I = y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x$

$= \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x} \cdot \frac{1}{\cos x} + \frac{\sin x}{\cos x} \cdot \sin^{2} x} d x = \int \frac{(1 + \sin^{2} x) \cos x}{(1 + \sin^{4} x)} d x$

Let $\sin x = t \Rightarrow \cos x d x = d t$

$∴ I = \int \frac{1 + t^{2}}{1 + t^{4}} d t = \int \frac{(1 + \frac{1}{t^{2}})}{(t^{2} + \frac{1}{t^{2}})} d t$

Put $t - \frac{1}{t} = u \Rightarrow (1 + \frac{1}{t^{2}}) d t = d u and t^{2} + \frac{1}{t^{2}} = u^{2} + 2$

$∴ I = \int \frac{d u}{u^{2} + 2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (t - \frac{1}{t}) + C$

$y (x) = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (\sin x - \cos e c x) + C$

$\lim_{x \to {(\frac{π}{2})}^{-}} y (x) = \lim_{h \to 0} [\frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} {\sin (\frac{π}{2} - h) - c o s e c (\frac{π}{2} - h)} + c]$

$\Rightarrow \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (0) + c = 0 \Rightarrow c = 0$

$∴ y (\frac{π}{4}) = \frac{1}{\sqrt{2}} \tan^{- 1} {\frac{1}{\sqrt{2}} (\sin \frac{π}{4} - c o s e c \frac{π}{4})}$

$= \frac{1}{\sqrt{2}} \tan^{- 1} {\frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} - \sqrt{2})} = \frac{1}{\sqrt{2}} \tan^{- 1} (- \frac{1}{2})$

Q 6 :

If $\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^{3} x \cos^{3} x \sin (x - θ)}} d x = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C,$ where C is the integration constant, then AB is equal to [2024]

$4 c o s e c (2 θ)$
$2 \sec θ$
$4 \sec θ$
$8 c o s e c (2 θ)$

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(4)

Let $I = \int \frac{\sin^{3 / 2} x + \cos^{3 / 2} x}{\sqrt{\sin^{3} x \cdot \cos^{3} x \cdot \sin (x - θ)}} d x$

$= \int \frac{d x}{\sqrt{\cos^{3} x \cdot \sin (x - θ)}} + \int \frac{d x}{\sqrt{\sin^{3} x \cdot \sin (x - θ)}}$

$= \int \frac{d x}{\cos^{2} x \sqrt{\frac{\sin x \cos θ - \cos x \sin θ}{\cos x}}} + \int \frac{d x}{\sin^{2} x \sqrt{\frac{\sin x \cos θ - \cos x \sin θ}{\sin x}}}$

$= \int \frac{\sec^{2} x}{\sqrt{\tan x \cos θ - \sin θ}} d x + \int \frac{c o s e c^{2} x}{\sqrt{\cos θ - \cot x \sin θ}} d x$

$\Rightarrow I = I_{1} + I_{2} (Let)$ ...(i)

Now, $I_{1} = \int \frac{\sec^{2} x d x}{\sqrt{\tan x \cos θ - \sin θ}}$

Put $\tan x \cos θ - \sin θ = t \Rightarrow \cos θ \cdot \sec^{2} x d x = d t$

$∴ I_{1} = \int \frac{d t}{\cos θ \cdot \sqrt{t}} \Rightarrow I_{1} = \frac{2 t^{1 / 2}}{\cos θ}$

$\Rightarrow I_{1} = \frac{2 {(\tan x \cos θ - \sin θ)}^{1 / 2}}{\cos θ}$

And, $I_{2} = \int \frac{c o s e c^{2} x}{\sqrt{\cos θ - \cot x \sin θ}} d x$

Put $\cos θ - \cot x \sin θ = v \Rightarrow \sin θ c o s e c^{2} x d x = d v$

$∴ I_{2} = \int \frac{d v}{\sin θ \sqrt{v}} = \frac{2 \sqrt{v}}{\sin θ} \Rightarrow I_{2} = \frac{2 \sqrt{\cos θ - \cot x \sin θ}}{\sin θ}$

From (i),

$I = \frac{2 \sqrt{\tan x \cos θ - \sin θ}}{\cos θ} + \frac{2 \sqrt{\cos θ - \cot x \sin θ}}{\sin θ} + C$ and

$I = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C$

So, $A = \frac{2}{\cos θ}, B = \frac{2}{\sin θ}$

$∴ A B = \frac{4 \cdot 2}{2 \sin θ \cos θ} = 8 c o s e c 2 θ$

Q 7 :

If $\int c o s e c^{5} x d x = α \cot x c o s e c x (c o s e c^{2} x + \frac{3}{2}) + β \log_{e} | \tan \frac{x}{2} | + C,$ where $α, β \in R$ and $C$ is the constant of integration, then the value of $8 (α + β)$ equals _______ . [2024]

(1)

$I_{n} = \int c o s e c^{n} x d x$

Using reduction formula, we get

$I_{n} = \frac{- c o s e c^{n - 2} x \cot x}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2}$

$∴$ $I_{5} = \int c o s e c^{5} x d x$

$= \frac{- c o s e c^{3} x \cot x}{4} + \frac{3}{4} I_{3} + C$

$= \frac{- c o s e c^{3} x \cot x}{4} + \frac{3}{4} [\frac{- c o s e c x \cot x}{2} + \frac{1}{2} \int c o s e c x d x] + C$

$= \frac{- 1}{4} c o s e c^{3} x \cot x - \frac{3}{8} c o s e c x \cot x + \frac{3}{8} \log_{e} | \tan \frac{x}{2} | + C$

$= \frac{- 1}{4} c o s e c x \cot x [c o s e c^{2} x + \frac{3}{2}] + \frac{3}{8} \log_{e} | \tan \frac{x}{2} | + C$

$= α \cot x c o s e c x [c o s e c^{2} x + \frac{3}{2}] + β \log_{e} | \tan \frac{x}{2} | + C (∵ Given)$

On comparing, we get $α = \frac{- 1}{4}$ and $β = \frac{3}{8}$

Hence, $8 (α + β) = 8 (\frac{- 1}{4} + \frac{3}{8}) = 8 \times \frac{1}{8} = 1$

Q 8 :

$If \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x = A {(\frac{α x - 1}{β x + 3})}^{B} + C, where C is the constant of integration,$ $then the value of α + β + 20 A B is ____ .$ [2024]

(7)

$I = \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x$

Putting $\frac{x + 3}{x - 1} = t \Rightarrow x = (\frac{3 + t}{t - 1}) \Rightarrow d x = \frac{- 4}{{(t - 1)}^{2}} d t$

$\Rightarrow {(x - 1)}^{4} {(x + 3)}^{6} = {(x - 1)}^{5} {(x + 3)}^{5} (\frac{x + 3}{x - 1})$

$∴ I = \int \frac{\frac{- 4}{{(t - 1)}^{2}} d t}{t^{1 / 5} (\frac{3 + t}{t - 1} - 1) (\frac{3 + t}{t - 1} + 3)}$

$I = \int \frac{- 4 d t}{t^{1 / 5} (16 t)} = \frac{- 1}{4} \int \frac{d t}{t^{6 / 5}} = \frac{- 1}{4} (\frac{t^{- 1 / 5}}{- 1 / 5}) + C$

On comparing, we get

$A = \frac{5}{4}, B = \frac{1}{5}, α = 1, β = 1$

Hence, $α + β + 20 A B = 1 + 1 + 20 \times \frac{5}{4} \times \frac{1}{5} = 7$

Q 9 :

Let $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$ . If $5 f (\sqrt{2}) = - 4$ , then f(1) is equal to [2025]

$- \frac{6 \sqrt{2}}{5}$
$- \frac{4 \sqrt{2}}{5}$
$- \frac{2 \sqrt{2}}{5}$
$- \frac{8 \sqrt{2}}{5}$

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(1)

We have, $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$

Put $3 - x^{2} = u^{2} \Rightarrow x d x = - u d u$

$∴ f (x) = \int (3 - u^{2}) \cdot u (- u d u)$

$= \int (u^{4} - 3 u^{2}) d u = \frac{u^{5}}{5} - u^{3} + C$

$\Rightarrow f (x) = \frac{{(3 - x^{2})}^{5 / 2}}{5} - {(3 - x^{2})}^{3 / 2} + C$ $[∵ u = \sqrt{3 - x^{2}}]$

$\Rightarrow f (\sqrt{2}) = \frac{1}{5} - 1 + C = \frac{- 4}{5} + C$

Now, $5 f (\sqrt{2}) = - 4$

$\Rightarrow - 4 + 5 C = - 4 \Rightarrow 5 C = 0 \Rightarrow C = 0$

Now, $f (1) = \frac{2^{5 / 2}}{5} - 2^{3 / 2} = 2^{1 / 2} (\frac{4}{5} - 2)$

$\Rightarrow f (1) = \frac{- 6 \sqrt{2}}{5}$ .

Q 10 :

If $\int e^{x} (\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{\sin^{- 1} x}{({(1 - x^{2})}^{3 / 2})} + \frac{x}{1 - x^{2}}) d x = g (x) + C$ , where C is the constant of integration, then $g (\frac{1}{2})$ equals: [2025]

$\frac{π}{6} \sqrt{\frac{e}{2}}$
$\frac{π}{6} \sqrt{\frac{e}{3}}$
$\frac{π}{4} \sqrt{\frac{e}{2}}$
$\frac{π}{4} \sqrt{\frac{e}{3}}$

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(2)

Let $I = \int e^{x} (\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{\sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} + \frac{x}{1 - x^{2}}) d x$

Now, $\frac{d}{d x} (\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}}) = \frac{\sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} + \frac{x}{1 - x^{2}}$

$\Rightarrow I = e^{x} \cdot \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + C$

$[∵ \int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + c]$

= g(x) + c [Given]

$∴ g (x) = \frac{x e^{x} \sin^{- 1} x}{\sqrt{1 - x^{2}}}$

Hence, $g (1 / 2) = \frac{e^{1 / 2}}{2} \cdot \frac{\frac{π}{6} \times 2}{\sqrt{3}} = \frac{π}{6} \sqrt{\frac{e}{3}}$ .

Q 11 :

Let $I (x)$ $= \int \frac{d x}{{(x - 11)}^{\frac{11}{13}} {(x + 15)}^{\frac{15}{13}}}$ . If $I (37) - I (24) = \frac{1}{4} (\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}})$ , $b, c \in N$ , then 3(b + c) is equal to [2025]

39
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40
26

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(1)

Given, $I (x)$ $= \int \frac{d x}{{(x - 11)}^{11 / 13} {(x + 15)}^{15 / 13}}$

Let $\frac{x - 11}{x + 15} = t \Rightarrow \frac{26 d x}{{(x + 15)}^{2}} = d t$

$∴ I (x)$ $= \frac{1}{26} \int \frac{d t}{{(t)}^{11 / 13}}$

$= \frac{1}{26} \times \frac{t^{2 / 13}}{\frac{2}{13}} + C$

$= \frac{1}{4} t^{2 / 13} + C$

$∴ I (x)$ $= \frac{1}{4} {(\frac{x - 11}{x + 15})}^{2 / 13} + C$

$∴ I (37) - I (24) = \frac{1}{4} [{(\frac{1}{2})}^{2 / 13} - {(\frac{1}{3})}^{2 / 13}]$

$\frac{1}{4} (\frac{1}{b^{1 / 13}} - \frac{1}{c^{1 / 13}}) = \frac{1}{4} (\frac{1}{{(4)}^{1 / 13}} - \frac{1}{9^{1 / 13}})$

$∴$ b = 4 and c = 9

$∴$ 3(b + c) = 3(4 + 9) = 39.

Q 12 :

Let $\int x^{3} \sin x d x = g (x) + C$ , where C is the constant of integration. If $8 (g (\frac{π}{2}) + g' (\frac{π}{2})) = α π^{3} + {βπ}^{2} + γ, α, β, γ \in Z$ , then $α + β - γ$ equals: [2025]

47
62
48
55

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(4)

$\int x^{3} \sin x d x = - x^{3} \cos x + \int 3 x^{2} \cos x d x$

$= - x^{3} \cos x + 3 x^{2} \sin x - \int 6 x^{} \sin x d x$

$= - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x + C$

So, $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 x \cos x - 6 \sin x$

and $g' (x) = - 3 x^{2} \cos x + x^{3} \sin x + 3 x^{2} \cos x + 6 x \sin x - 6 x \sin x + 6 \cos x - 6 \cos x$

$= x^{3} \sin x$

$\Rightarrow g (\frac{π}{2}) = \frac{3 π^{2}}{4} - 6 and g' (\frac{π}{2}) = \frac{π^{3}}{8}$

$∴ 8 (g (\frac{π}{2}) + g' (\frac{π}{2})) = π^{3} + 6 π^{2} - 48$

$\Rightarrow α = 1, β = 6, γ = - 48$

So, $α + β - γ = 55$ .

Q 13 :

Let $f : (0, \infty) \to R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f' (x) = 2 x f (x) + 3$ , with f(1) = 4. Then 2f(2) is equal to : [2025]

23
19
29
39

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(4)

Given, $x^{2} f' (x) = 2 x f (x) + 3$ and f(1) = 4

$\Rightarrow x^{2} f' (x) - 2 x f (x) = 3$

$\Rightarrow \frac{x^{2} f' (x) - 2 x f (x)}{x^{4}} = \frac{3}{x^{4}}$ (Divide both sides by $x^{4}$ )

$\Rightarrow \frac{d}{d x} (\frac{f (x)}{x^{2}}) = \frac{3}{x^{4}}$

Using integration on both sides,

$\Rightarrow \frac{f (x)}{x^{2}} = \int \frac{3}{x^{4}} d x$

$\Rightarrow \frac{f (x)}{x^{2}} = 3 \times \frac{- 1}{3 x^{3}} + C$

$\Rightarrow \frac{f (x)}{x^{2}} = \frac{- 1}{x^{3}} + C$

$\Rightarrow f (x) = \frac{- 1}{x} + C x^{2}$

Since f(1) = 4

$\Rightarrow 4 = - 1 + C \Rightarrow C = 4 + 1 = 5$

Now, we get f(x) $= \frac{- 1}{x} + 5 x^{2}$

$\Rightarrow 2 f (2) = 2 \cdot (\frac{- 1}{2} + 5 \times 4)$

$= 2 \times \frac{39}{2} = 39$ .

Q 14 :

If $f (x) = \int \frac{1}{x^{1 / 4} (1 + x^{1 / 4})} d x$ , f(0) = – 6, then f(1) is equal to : [2025]

$\log_{e} 2 + 2$
$4 (\log_{e} 2 - 2)$
$4 (\log_{e} 2 + 2)$
$2 - \log_{e} 2$

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(2)

We have, $f (x) = \int \frac{1}{x^{1 / 4} (1 + x^{1 / 4})} d x$

Putting $x = t^{4} \Rightarrow d x = 4 t^{3} d t$

$f (t^{4}) = \int \frac{4 t^{3}}{t (1 + t)} d t = 4 \int \frac{t^{2}}{1 + t} d t$

$= 4 \int \frac{(t^{2} - 1) + 1}{1 + t} d t$

$= 4 [\int (t - 1) d t + \int \frac{1}{t + 1} d t]$

$= 4 [\frac{t^{2}}{2} - t + \log_{e} (t + 1)] + C$

$∴ f (x) = 2 x^{1 / 2} - 4 x^{1 / 4} + 4 \log_{e} (x^{1 / 4} + 1) + C$

$f (0) = - 6 \Rightarrow C = - 6$

$∴ f (1) = 2 - 4 + 4 \log_{e} 2 - 6 = 4 \log_{e} 2 - 8 = 4 (\log_{e} 2 - 2)$

Q 15 :

If $\int \frac{{(\sqrt{1 + x^{2}} + x)}^{10}}{{(\sqrt{1 + x^{2}} - x)}^{9}} d x = \frac{1}{m} ({(\sqrt{1 + x^{2}} + x)}^{n} (n \sqrt{1 + x^{2}} - x)) + C$ where C is the constant of integration and m, n $\in$ N, then m + n is equal to __________. [2025]

(379)

Let $I = \int \frac{{(\sqrt{1 + x^{2}} + x)}^{10}}{{(\sqrt{1 + x^{2}} - x)}^{9}} d x = \int \frac{{(\sqrt{1 + x^{2}} + x)}^{19}}{1} d x$ (On rationalising)

Let $\sqrt{1 + x^{2}} + x = t$

$\Rightarrow (\frac{x}{\sqrt{1 + x^{2}}} + 1) d x = d t$

$\Rightarrow d x = \frac{d t}{t} (\sqrt{1 + x^{2}})$

$= \frac{d t}{t} \cdot (\frac{t^{2} + 1}{2 t}) = \frac{t^{2} + 1}{2 t^{2}} \cdot d t$

So, $I = \int t^{19} (\frac{t^{2} + 1}{2 t^{2}}) d t$

$= \frac{1}{2} \int (t^{19} + t^{17}) d t = \frac{1}{2} [\frac{t^{20}}{20} + \frac{t^{18}}{18}] + C$

$= \frac{t^{19}}{360} [9 t + \frac{10}{t}] + C = \frac{t^{19}}{360} [9 (t + \frac{1}{t}) + \frac{1}{t}]$

$= \frac{{(\sqrt{1 + x^{2}} + x)}^{19}}{360} [9 (2 \sqrt{1 + x^{2}}) + (\sqrt{1 + x^{2}} - x)] + C$

$= \frac{{(\sqrt{1 + x^{2}} + x)}^{19}}{360} [19 \sqrt{1 + x^{2}} - x] + C$

Then, m = 360, n = 19

$∴$ m + n = 360 + 19 = 379.

Q 16 :

If $\int (\frac{1}{x} + \frac{1}{x^{3}}) (\sqrt[23]{3 x^{- 24} + x^{- 26}}) d x = - \frac{α}{3 (α + 1)} {(3 x^{β} + x^{γ})}^{\frac{α + 1}{α}} + C$ , x > 0, $(α, β, γ \in Z)$ , where c is the constant of integration, then $α + β + γ$ is equal to __________. [2025]

(19)

We have, $\int (\frac{1}{x^{2}} + \frac{1}{x^{4}}) {(\frac{3}{x} + \frac{1}{x^{3}})}^{\frac{1}{23}} d x$

Put $t = \frac{3}{x} + \frac{1}{x^{3}} \Rightarrow d t = - 3 (\frac{1}{x^{2}} + \frac{1}{x^{4}}) d x$

Now, $\int \frac{t^{1 / 23} d t}{- 3} = \frac{t^{24 / 23}}{(\frac{24}{23}) (- 3)} + C$

On comparing, we get $α = 23, β = - 1, γ = - 3$

$∴ α + β + γ = 19$ .

Q 17 :

If $\int \frac{2 x^{2} + 5 x + 9}{\sqrt{x^{2} + x + 1}} d x = x \sqrt{x^{2} + x + 1} + α \sqrt{x^{2} + x + 1} + β \log_{e}$ $| x + \frac{1}{2} + \sqrt{x^{2} + x + 1} |$ $+ C$ , where C is the constant of integration, then $α + 2 β$ is equal to __________. [2025]

(16)

Given integral is $\int \frac{2 x^{2} + 5 x + 9}{\sqrt{x^{2} + x + 1}} d x$

Using partial fraction decomposition, we get

$2 x^{2} + 5 x + 9 = A (x^{2} + x + 1) + B (2 x + 1) + C$

$= A x^{2} + (A + 2 B) x + A + B + C$

Comparing terms, we get $A = 2, B = \frac{3}{2} and C = \frac{11}{2}$

$∴ \int \frac{2 x^{2} + 5 x + 9}{\sqrt{x^{2} + x + 1}} d x = 2 \int \frac{x^{2} + x + 1}{\sqrt{x^{2} + x + 1}} d x + \frac{3}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x + 1}} d x + \frac{11}{2} \int \frac{1}{\sqrt{x^{2} + x + 1}} d x$

$= 2 \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x + \frac{3}{2} \times 2 \sqrt{x^{2} + x + 1} + \frac{11}{2} \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} d x$

$= 2 \times ((\frac{x + \frac{1}{2}}{2}) \sqrt{x^{2} + x + 1} + \frac{3}{2 \times 4} \ln | x + \frac{1}{2} + \sqrt{x^{2} + x + 1} |) + 3 \sqrt{x^{2} + x + 1} + \frac{11}{2} \ln | x + \frac{1}{2} + \sqrt{x^{2} + x + 1} | + C$

On comparing, we get $α = \frac{7}{2} and β = \frac{25}{4}$

Hence, $α + 2 β = \frac{7}{2} + 2 \times \frac{25}{4} = 16$ .

Q 18 :

$Let I (x) = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x . If I (0) = 0, then I (\frac{π}{4}) is equal to$ [2023]

$\log_{e} \frac{{(π + 4)}^{2}}{16} + \frac{π^{2}}{4 (π + 4)}$
$\log_{e} \frac{{(π + 4)}^{2}}{32} + \frac{π^{2}}{4 (π + 4)}$
$\log_{e} \frac{{(π + 4)}^{2}}{16} - \frac{π^{2}}{4 (π + 4)}$
$\log_{e} \frac{{(π + 4)}^{2}}{32} - \frac{π^{2}}{4 (π + 4)}$

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(4)

$I (x) = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x$

By using integration by parts

$I (x) = x^{2} \int \frac{(x \sec^{2} x + \tan x) d x}{{(x \tan x + 1)}^{2}} - \int (2 x) \int \frac{(x \sec^{2} x + \tan x) d x}{{(x \tan x + 1)}^{2}}$

$I (x) = x^{2} \frac{(- 1)}{(x \tan x + 1)} + \int \frac{2 x}{x \tan x + 1} d x$

$I (x) = x^{2} (\frac{- 1}{x \tan x + 1}) + 2 \int \frac{x \cos x}{x \sin x + \cos x} d x$

$I (x) = x^{2} (\frac{- 1}{x \tan x + 1}) + 2 \ln$ $| x \sin x + \cos x |$ $+ C$

$As, I (0) = 0$

$\Rightarrow 0 = 0 + 2 \ln | 1 | + C$ $∴$ $C = 0$

$∴$ $I (\frac{π}{4}) = \frac{- π^{2}}{16} (\frac{1}{\frac{π}{4} + 1}) + 2 \ln$ $| \frac{π}{4} \sin (\frac{π}{4}) + \cos \frac{π}{4} |$

$= 2 \ln (\frac{(π + 4)}{4 \sqrt{2}}) - \frac{π^{2}}{4 (π + 4)} = \ln [\frac{{(π + 4)}^{2}}{32}] - \frac{π^{2}}{4 (π + 4)}$

Q 19 :

$Let I (x) = \int \frac{(x + 1)}{x {(1 + x e^{x})}^{2}} d x, x > 0 . If \lim_{x \to \infty} I (x) = 0, then I (1) is equal to$ [2023]

$\frac{e + 1}{e + 2} - \log_{e} (e + 1)$
$\frac{e + 2}{e + 1} - \log_{e} (e + 1)$
$\frac{e + 2}{e + 1} + \log_{e} (e + 1)$
$\frac{e + 1}{e + 2} + \log_{e} (e + 1)$

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(2)

$We have,$ $I (x) = \int \frac{(x + 1)}{x {(1 + x e^{x})}^{2}} d x$

$= \int \frac{e^{x} (1 + x)}{x e^{x} {(1 + x e^{x})}^{2}} d x$

$Put 1 + x e^{x} = t, we get$

$\Rightarrow (x e^{x} + e^{x}) d x = d t \Rightarrow (1 + x) e^{x} d x = d t$

$∴$ $I (x) = \int \frac{d t}{(t - 1) t^{2}} = \int (- \frac{1}{t} - \frac{1}{t^{2}} + \frac{1}{t - 1}) d t$

$= - \log t + \frac{1}{t} + \log (t - 1) + c$

$= - \log (1 + x e^{x}) + \frac{1}{1 + x e^{x}} + \log (x e^{x}) + c$

$\Rightarrow I (x) = \frac{1}{1 + x e^{x}} + \log (\frac{x e^{x}}{1 + x e^{x}}) + c$

$Now, \lim_{x \to \infty} I (x) = 0 \Rightarrow c = 0$

$∴$ $I (1) = \frac{1}{1 + e} + \log (\frac{e}{1 + e})$

$= \frac{1}{1 + e} + 1 - \log (1 + e) \Rightarrow \frac{2 + e}{1 + e} - \log (1 + e)$

Q 20 :

$For α, β, γ, δ \in ℕ, if \int [{(\frac{x}{e})}^{2 x} + {(\frac{e}{x})}^{2 x}] \log_{e} x d x$ $= \frac{1}{α} {(\frac{x}{e})}^{β x} - \frac{1}{γ} {(\frac{e}{x})}^{δ x} + C, where e = \sum_{n = 0}^{\infty} \frac{1}{n!}$ $and C is constant of integration, then α + 2 β + 3 γ - 4 δ is equal to$ [2023]

1
- 8
4
- 4

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(3)

$We have,$ $\int [{(\frac{x}{e})}^{2 x} + {(\frac{e}{x})}^{2 x}] \log_{e} x d x$

$Putting {(\frac{x}{e})}^{2 x} = t$ $\Rightarrow 2 x (\log x - 1) = \log t$

$\Rightarrow [2 (\log x - 1) + 2] d x = \frac{1}{t} d t \Rightarrow 2 \log x d x = \frac{1}{t} d t,$ we get

$\int (t + \frac{1}{t}) \frac{1}{2 t} d t = \frac{1}{2} \int d t + \frac{1}{2} \int \frac{1}{t^{2}} d t$

$= \frac{1}{2} t - \frac{1}{2 t} + c$ $= \frac{1}{2} [{(\frac{x}{e})}^{2 x} - {(\frac{e}{x})}^{2 x}] + c$

$Comparing with \frac{1}{α} {(\frac{x}{e})}^{β x} - \frac{1}{γ} {(\frac{e}{x})}^{δ x} + c$

$We get, α = 2, β = 2, γ = 2, δ = 2$

$Now, α + 2 β + 3 γ - 4 δ = 2 + 4 + 6 - 8 = 4$

Q 21 :

$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

$\log_{e} 17 - \log_{e} 18$
$\log_{e} 19 - \log_{e} 20$
$\frac{1}{2} (\log_{e} 19 - \log_{e} 17)$
$\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

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(4)

$Let x^{2} = t, 2 x d x = d x$

$f (x) = \int \frac{d t}{(t + 1) (t + 3)} = \frac{1}{2} \int (\frac{1}{t + 1} - \frac{1}{t + 3}) d t$

$f (x) = \frac{1}{2} \log_{e} (\frac{t + 1}{t + 3}) + c = \frac{1}{2} \log_{e} (\frac{x^{2} + 1}{x^{2} + 3}) + c$

$f (3) = \frac{1}{2} \log_{e} (\frac{10}{12}) + c = \frac{1}{2} \log_{e} \frac{5}{6} + c = \frac{1}{2} (\log_{e} 5 - \log_{e} 6) \Rightarrow c = 0$

$f (x) = \frac{1}{2} \log_{e} (\frac{x^{2} + 1}{x^{2} + 3}); f (4) = \frac{1}{2} \log_{e} (\frac{17}{19}) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

Q 22 :

Let $I (x) = \int \sqrt{\frac{x + 7}{x}} d x$ and $I (9) = 12 + 7 \log_{e} 7$ . If $I (1) = α + 7 \log_{e} (1 + 2 \sqrt{2})$ then $α^{4}$ is equal to _______ . [2023]

(64)

$We have, I (x) = \int \sqrt{\frac{x + 7}{x}} d x$

$Put x = t^{2} \Rightarrow d x = 2 t d t$

$So, \int 2 \sqrt{t^{2} + 7} d t = 2 \int \sqrt{t^{2} + {(\sqrt{7})}^{2}} d t$

$= 2 [\frac{t}{2} \sqrt{t^{2} + 7} + \frac{7}{2} \ln | t + \sqrt{t^{2} + 7} |] + C$

$\Rightarrow I (x) = \sqrt{x} \sqrt{x + 7} + 7 \ln$ $| \sqrt{x} + \sqrt{x + 7} |$ $+ C$

$We have, I (9) = 12 + 7 \ln 7 = 12 + 7 [\ln (3 + 4)]$

$\Rightarrow C = 0$

$So, I (x) = \sqrt{x} \sqrt{x + 7} + 7 \ln$ $| \sqrt{x} + \sqrt{x + 7} |$

$I (1) = \sqrt{8} + 7 \ln$ $| 1 + \sqrt{8} |$

$∴$ $I (1) = \sqrt{8} + 7 \ln$ $| 1 + 2 \sqrt{2} |$ $\Rightarrow α = \sqrt{8}$

$∴$ $α^{4} = {[{(8)}^{1 / 2}]}^{4} = 8^{2} \Rightarrow α^{4} = 64$

Q 23 :

Let $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}},$ $| x |$ $< \frac{2}{\sqrt{3}}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , $α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______ . [2023]

(28)

$Given, f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}}$

$x = \frac{1}{t} \Rightarrow d x = \frac{- 1}{t^{2}} d t$

$\Rightarrow f (x) = \int \frac{- \frac{1}{t^{2}} d t}{\frac{(3 t^{2} + 4)}{t^{2}} \frac{\sqrt{4 t^{2} - 3}}{t}}$

$= - \int \frac{- t d t}{(3 t^{2} + 4) \sqrt{4 t^{2} - 3}}$ $Put 4 t^{2} - 3 = z^{2} \Rightarrow t d t = \frac{z}{4} d z$

$=$ $- \frac{1}{4} \int \frac{z d z}{(3 (\frac{z^{2} + 3}{4}) + 4) z} = \int \frac{- d z}{3 z^{2} + 25} = - \frac{1}{3} \int \frac{d z}{z^{2} + {(\frac{5}{\sqrt{3}})}^{2}}$

$= \int - \frac{1}{3} \frac{\sqrt{3}}{5} \tan^{- 1} (\frac{\sqrt{3} z}{5}) + C = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5} \sqrt{4 t^{2} - 3}) + C$

$f (x) = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5} \sqrt{\frac{4 - 3 x^{2}}{x^{2}}}) + C$

$∵$ $f (0) = 0$ $∴$ $C = \frac{π}{10 \sqrt{3}}$

$Now, f (1) = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5}) + \frac{π}{10 \sqrt{3}}$

$\Rightarrow f (1) = \frac{1}{5 \sqrt{3}} \cot^{- 1} (\frac{\sqrt{3}}{5}) = \frac{1}{5 \sqrt{3}} \tan^{- 1} \frac{5}{\sqrt{3}}$

$So, α = 5 : β = \sqrt{3}$ $∴$ $α^{2} + β^{2} = 28$

Q 24 :

If $\int \sqrt{\sec 2 x - 1} d x = α \log_{e}$ $| \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} |$ $+$ constant, then $β - α$ is equal to _________ . [2023]

(1)

$Let I = \int \sqrt{(\sec 2 x - 1)} d x = \int \sqrt{\frac{1 - \cos 2 x}{\cos 2 x}} d x$

$= \int \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x - 1}} d x$ $= \sqrt{2} \int \frac{\sin x}{\sqrt{2 \cos^{2} x - 1}} d x$

$Substitute \cos x = t \Rightarrow - \sin x d x = d t$

$I = - \sqrt{2} \int \frac{d t}{\sqrt{2 t^{2} - 1}} = - \ln$ $| \sqrt{2} t + \sqrt{2 t^{2} - 1} |$

$= - \ln$ $| \sqrt{2} \cos x + \sqrt{2 \cos^{2} x - 1} |$

$= - \ln$ $| \sqrt{2} \cos x + \sqrt{\cos 2 x} |$

$= - \frac{1}{2} \ln$ $| 2 \cos^{2} x + \cos 2 x + 2 \sqrt{\cos 2 x} \cdot \sqrt{2} \cos x |$ $+ C$

$= \frac{- 1}{2} \ln (2 \cos 2 x + 1 + 2 \sqrt{\cos 2 x} \sqrt{1 + \cos 2 x})$

$= \frac{- 1}{2} \ln (\cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x} \sqrt{1 + \cos 2 x})$

$= \frac{- 1}{2} \ln$ $| \cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x} \cdot \sqrt{1 + \cos 2 x} |$ $+ C$

$\Rightarrow - \frac{1}{2} \ln$ $| \cos x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)} |$ $+ C$

$= α \ln$ $| \cos 2 x + β |$ $+ \sqrt{\cos 2 x (1 + \cos \frac{1}{β})} + c$

On comparing, we get, $α = - \frac{1}{2}, β = \frac{1}{2}$ $∴$ $β - α = \frac{1}{2} + \frac{1}{2} = 1$

Q 25 :

Let $f (x)$ $= \int \frac{(2 - x^{2}) . e^{x}}{(\sqrt{1 + x}) {(1 - x)}^{3 / 2}} d x .$ If $f (0) = 0$ , then $f (\frac{1}{2})$ is equal to: [2026]

$\sqrt{3 e} + 1$
$\sqrt{2 e} - 1$
$\sqrt{2 e} + 1$
$\sqrt{3 e} - 1$

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(4)

$\int e^{x} (\frac{(1 - x^{2}) + 1}{\sqrt{1 + x} . {(1 - x)}^{3 / 2}}) d x$

$= \int e^{x} (\frac{(1 - x^{2})}{\sqrt{1 + x} . {(1 - x)}^{3 / 2}} + \frac{1}{\sqrt{1 + x} . {(1 - x)}^{3 / 2}}) d x$

$= \int e^{x} (\sqrt{\frac{1 + x}{1 - x}} + \frac{1}{\sqrt{1 + x} . {(1 - x)}^{3 / 2}}) d x$

$= e^{x} \sqrt{\frac{1 + x}{1 - x}} + C$

$f (x) = e^{x} \sqrt{\frac{1 + x}{1 - x}} - 1$

$f (\frac{1}{2}) = \sqrt{3 e} - 1$

Q 26 :

Let $f (t)$ $= \int (\frac{1 - \sin (\log_{e} t)}{1 - \cos (\log_{e} t)}) d t, t > 1 .$

If $f (e^{π / 2}) = - e^{π / 2} and f (e^{π / 4}) = α e^{π / 4},$ then $α$ equals [2026]

$- 1 - \sqrt{2}$
$1 + \sqrt{2}$
$- 1 - 2 \sqrt{2}$
$- 1 + \sqrt{2}$

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(1)

$f (t) = \int \frac{1 - \sin (\ln t)}{1 - \cos (\ln t)} d t$

$Let \ln t = x \Rightarrow t = e^{x} \Rightarrow d t = e^{x} d x$

$= \frac{1}{2} \int (\cos e c^{2} \frac{x}{2} - 2 \cot \frac{x}{2}) e^{x} d x - t \cot (\frac{\ln t}{2}) + C$

$(∴ \int (f (x) + f' (x)) e^{x} d x = f (x) . e^{x} + C)$

$Now f (e^{π / 2}) = - e^{π / 2} \cot (\frac{π}{4}) + C = - e^{π / 2} (given)$

$\Rightarrow C = 0$

$Now f (e^{π / 4}) = - e^{π / 4} \cot (\frac{π}{8}) + C = - e^{π / 4} (\sqrt{2} + 1)$

Q 27 :

Let $f (x)$ $= \int \frac{d x}{x^{(\frac{2}{3})} + 2 x^{(\frac{1}{2})}}$ be such that $f (0) = - 26 + 24 \log_{e} (2) .$ If $f (1) = a + b \log_{e} (3),$ where $a, b \in ℤ$ , then a+b is equal to: [2026]

-11
-26
-18
-5

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(1)

$f (x) = \int \frac{d x}{x^{2 / 3} + 2 x^{1 / 2}}$

$Put x = t^{6} \Rightarrow d x = 6 t^{5} d t$

$= \int \frac{6 t^{5}}{t^{4} + 2 t^{3}} d t = 6 \int \frac{(t^{2} - 4) + 4}{t + 2} d t$

$= 6 [\int (t - 2) d t + 4 \int \frac{1}{t + 2} d t]$

$= 6 [\frac{t^{2}}{2} - 2 t + 4 l n (t + 2)] + C$

$= 3 x^{1 / 3} - 12 x^{1 / 6} + 24 l n (x^{1 / 6} + 2) + C$

$f (0) = 24 l n 2 + C = - 26 + 24 l n 2 (given)$

$\Rightarrow C = - 26$

$Now$

$f (1) = - 35 + 24 l n 3 = a + b l n 3 (as given in ques.)$

$\Rightarrow a = - 35, b = 24$

$\Rightarrow a + b = - 11$

Q 28 :

If $\int {(\sin x)}^{\frac{- 11}{2}} {(\cos x)}^{\frac{- 5}{2}} d x = - \frac{p_{1}}{q_{1}} {(\cot x)}^{\frac{9}{2}} -$ $\frac{p_{2}}{q_{2}} {(\cot x)}^{\frac{5}{2}} - \frac{p_{3}}{q_{3}} {(\cot x)}^{\frac{1}{2}} +$ $\frac{p_{4}}{q_{4}} {(\cot x)}^{\frac{- 3}{2}} + C$

where $p_{i}$ and $q_{i}$ are positive integers with $\gcd (p_{i}, q_{i}) = 1$ for $i = 1, 2, 3, 4$ , and C is the constant of integration, then $\frac{15 p_{1} p_{2} p_{3} p_{4}}{q_{1} q_{2} q_{3} q_{4}}$ is equal to ______ [2026]

(16)

$\int {(\tan x)}^{- 11 / 2} . \sec^{8} x d x$

$= \int {(\tan x)}^{- 11 / 2} {(1 + \tan^{2} x)}^{3} \sec^{2} x d x$

Put $\tan x = t$

$\Rightarrow \int t^{- 11 / 2} {(1 + t^{2})}^{3} d t$ $= \int t^{- 11 / 2} (1 + 3 t^{2} + 3 t^{4} + t^{6}) d t$

$= \int (t^{- 11 / 2} + 3 t^{- 7 / 2} + 3 t^{- 3 / 2} + t^{1 / 2}) d t$

$= - \frac{2}{9} {(\cot x)}^{9 / 2} - \frac{6}{5} {(\cot x)}^{5 / 2} - 6 {(\cot x)}^{1 / 2} + \frac{2}{3} {(\cot x)}^{- 3 / 2} + C$

$\Rightarrow p_{1} = 2, p_{2} = 6, p_{3} = 6, p_{4} = 2$

$and q_{1} = 9, q_{2} = 5, q_{3} = 1, q_{4} = 3$

$\frac{15 p_{1} p_{2} p_{3} p_{4}}{q_{1} q_{2} q_{3} q_{4}} = \frac{15 \cdot 2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 16$

Q 29 :

Let f be a differentiable function satisfying

$f (x) = 1 - 2 x + \int_{0}^{x} e^{(x - t)} f (t) d t, x \in ℝ$ and let $g (x) = \int_{0}^{x} (f {(t) + 2)}^{15} {(t - 4)}^{6} {(t + 12)}^{17} d t, x \in ℝ .$

If p and q are respectively the points of local minima and local maxima of g, then the value of $| p + q |$ is equal to ________ [2026]

(9)

$f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$e^{- x} f (x) = (1 - 2 x) e^{- x} + \int_{0}^{x} e^{- t} f (t) d t$

$e^{- x} f' (x) - e^{- x} f (x) = - 2 e^{- x} + (1 - 2 x) e^{- x} (- 1) + e^{- x} f (x)$

$f' (x) - 2 f (x) = 2 x - 3$

$\frac{d y}{d x} - 2 y = 2 x - 3$

$\Rightarrow y e^{- 2 x} = \int e^{- 2 x} (2 x - 3) d x$

$On solving we get f (x) = 1 - x$

$g (x) = \int_{0}^{x} {(3 - t)}^{15} {(t - 4)}^{6} {(t + 12)}^{17} d t$

$g' (x) = {(3 - x)}^{15} {(x - 4)}^{6} {(x + 12)}^{17}$

$= - {(x - 3)}^{15} {(x - 4)}^{6} {(x + 12)}^{17}$

$Local maxima \Rightarrow q = 3$

$Local minima \Rightarrow p = - 12$ $= | p + q | = 9$

Q 30 :

Let $f (x) = \int \frac{7 x^{10} + 9 x^{8}}{{(1 + x^{2} + 2 x^{9})}^{2}} d x, x > 0,$ $\lim_{x \to 0} f (x) = 0$ and $f (1) = \frac{1}{4}$ .

If $A = [\begin{matrix} 0 & 0 & 1 \\ \frac{1}{4} & f' (1) & 1 \\ α^{2} & 4 & 1 \end{matrix}]$ and $B = adj (adj A)$ be such that $| B |$ $= 81$ , then $α^{2}$ is equal to [2026]

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(2)

$f (x) = \int \frac{(\frac{7}{x^{8}} + \frac{9}{x^{10}})}{{(\frac{1}{x^{9}} + \frac{1}{x^{7}} + 2)}^{2}} d x$

$Put t = \frac{1}{x^{9}} + \frac{1}{x^{7}} + 2 \Rightarrow \frac{d t}{d x} = \frac{- 9}{x^{10}} - \frac{7}{x^{8}}$

$f (x) = \int \frac{- d t}{t^{2}} = \frac{1}{t} + C$

$f (x) = \frac{1}{\frac{1}{x^{9}} + \frac{1}{x^{7}} + 2} + C$

$= \frac{x^{9}}{1 + x^{2} + 2 x^{9}} + C$

$Given f (1) = \frac{1}{4} = \frac{1}{4} + C \Rightarrow C = 0$

$f (x) = \frac{x^{9}}{1 + x^{2} + 2 x^{9}}$

$f' (x) = \frac{(1 + x^{2} + 2 x^{9}) - 9 x^{8} - x^{9} (2 x + 18 x^{8})}{{(1 + x^{2} + 2 x^{9})}^{2}}$

$f' (x) = \frac{36 - 20}{16} = 1$

$A = (\begin{matrix} 0 & 0 & 1 \\ 4 & 1 & 1 \\ α^{2} & \frac{1}{4} & 1 \end{matrix})$

$| A |$ $=$ $| 1 - α^{2} |$ $= 3$

$1 - α^{2} = 3, - 3 \Rightarrow α^{2} = - 2, 4$

$Value of α^{2} = 4$

$B = adj (adj A)$

$| B |$ $= 81 =$ $| A |^{4}$ $\Rightarrow$ $| A |$ $= 3$

Topic Question Set

Q 1 :

$If \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + constant,$ $then the maximum value of a \sin x + b \cos x is:$ [2024]

Q 2 :

Let $I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x .$ If $I (0) = 3,$ then $I (\frac{π}{12})$ is equal to [2024]

Q 3 :

Let $\int \frac{2 - \tan x}{3 + \tan x} d x = \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C,$ where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to: [2024]

Q 4 :

The integral $\int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$ is equal to: [2024]

Q 5 :

$For x \in (- \frac{π}{2}, \frac{π}{2}), if y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x,$ $and \lim_{x \to {(\frac{π}{2})}^{-}} y (x) = 0, then y (\frac{π}{4}) is equal to$ [2024]

Q 6 :

If $\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^{3} x \cos^{3} x \sin (x - θ)}} d x = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C,$ where C is the integration constant, then AB is equal to [2024]

Q 7 :

If $\int c o s e c^{5} x d x = α \cot x c o s e c x (c o s e c^{2} x + \frac{3}{2}) + β \log_{e} | \tan \frac{x}{2} | + C,$ where $α, β \in R$ and $C$ is the constant of integration, then the value of $8 (α + β)$ equals _______ . [2024]

Q 8 :

$If \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x = A {(\frac{α x - 1}{β x + 3})}^{B} + C, where C is the constant of integration,$ $then the value of α + β + 20 A B is ____ .$ [2024]

Q 9 :

Let $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$ . If $5 f (\sqrt{2}) = - 4$ , then f(1) is equal to [2025]

Q 10 :

If $\int e^{x} (\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{\sin^{- 1} x}{({(1 - x^{2})}^{3 / 2})} + \frac{x}{1 - x^{2}}) d x = g (x) + C$ , where C is the constant of integration, then $g (\frac{1}{2})$ equals: [2025]

Q 11 :

Let $I (x)$ $= \int \frac{d x}{{(x - 11)}^{\frac{11}{13}} {(x + 15)}^{\frac{15}{13}}}$ . If $I (37) - I (24) = \frac{1}{4} (\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}})$ , $b, c \in N$ , then 3(b + c) is equal to [2025]

Q 12 :

Let $\int x^{3} \sin x d x = g (x) + C$ , where C is the constant of integration. If $8 (g (\frac{π}{2}) + g' (\frac{π}{2})) = α π^{3} + {βπ}^{2} + γ, α, β, γ \in Z$ , then $α + β - γ$ equals: [2025]

Q 13 :

Let $f : (0, \infty) \to R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f' (x) = 2 x f (x) + 3$ , with f(1) = 4. Then 2f(2) is equal to : [2025]

Q 14 :

If $f (x) = \int \frac{1}{x^{1 / 4} (1 + x^{1 / 4})} d x$ , f(0) = – 6, then f(1) is equal to : [2025]

Q 15 :

If $\int \frac{{(\sqrt{1 + x^{2}} + x)}^{10}}{{(\sqrt{1 + x^{2}} - x)}^{9}} d x = \frac{1}{m} ({(\sqrt{1 + x^{2}} + x)}^{n} (n \sqrt{1 + x^{2}} - x)) + C$ where C is the constant of integration and m, n $\in$ N, then m + n is equal to __________. [2025]

Q 16 :

If $\int (\frac{1}{x} + \frac{1}{x^{3}}) (\sqrt[23]{3 x^{- 24} + x^{- 26}}) d x = - \frac{α}{3 (α + 1)} {(3 x^{β} + x^{γ})}^{\frac{α + 1}{α}} + C$ , x > 0, $(α, β, γ \in Z)$ , where c is the constant of integration, then $α + β + γ$ is equal to __________. [2025]

Q 17 :

If $\int \frac{2 x^{2} + 5 x + 9}{\sqrt{x^{2} + x + 1}} d x = x \sqrt{x^{2} + x + 1} + α \sqrt{x^{2} + x + 1} + β \log_{e}$ $| x + \frac{1}{2} + \sqrt{x^{2} + x + 1} |$ $+ C$ , where C is the constant of integration, then $α + 2 β$ is equal to __________. [2025]

Q 18 :

$Let I (x) = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x . If I (0) = 0, then I (\frac{π}{4}) is equal to$ [2023]

Q 19 :

$Let I (x) = \int \frac{(x + 1)}{x {(1 + x e^{x})}^{2}} d x, x > 0 . If \lim_{x \to \infty} I (x) = 0, then I (1) is equal to$ [2023]

Q 21 :

$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

Q 22 :

Let $I (x) = \int \sqrt{\frac{x + 7}{x}} d x$ and $I (9) = 12 + 7 \log_{e} 7$ . If $I (1) = α + 7 \log_{e} (1 + 2 \sqrt{2})$ then $α^{4}$ is equal to _______ . [2023]

Q 23 :

Let $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}},$ $| x |$ $< \frac{2}{\sqrt{3}}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , $α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______ . [2023]

Q 24 :

If $\int \sqrt{\sec 2 x - 1} d x = α \log_{e}$ $| \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} |$ $+$ constant, then $β - α$ is equal to _________ . [2023]

Q 25 :

Let $f (x)$ $= \int \frac{(2 - x^{2}) . e^{x}}{(\sqrt{1 + x}) {(1 - x)}^{3 / 2}} d x .$ If $f (0) = 0$ , then $f (\frac{1}{2})$ is equal to: [2026]

Q 26 :

Let $f (t)$ $= \int (\frac{1 - \sin (\log_{e} t)}{1 - \cos (\log_{e} t)}) d t, t > 1 .$

If $f (e^{π / 2}) = - e^{π / 2} and f (e^{π / 4}) = α e^{π / 4},$ then $α$ equals [2026]

Q 27 :

Let $f (x)$ $= \int \frac{d x}{x^{(\frac{2}{3})} + 2 x^{(\frac{1}{2})}}$ be such that $f (0) = - 26 + 24 \log_{e} (2) .$ If $f (1) = a + b \log_{e} (3),$ where $a, b \in ℤ$ , then a+b is equal to: [2026]

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Topic Question Set

Q 1 : If ∫1a2sin2x+b2cos2x dx=112tan-1(3tanx)+constant, then the maximum value of asinx+bcosx is: [2024]

Q 2 : Let I(x)=∫6sin2x(1-cotx)2 dx. If I(0)=3, then I(π12) is equal to [2024]

Q 3 : Let ∫2-tanx3+tanx dx=12(αx+loge|βsinx+γcosx|)+C, where C is the constant of integration. Then α+γβ is equal to: [2024]

Q 4 : The integral ∫(x8-x2)dx(x12+3x6+1)tan-1(x3+1x3) is equal to: [2024]

Q 5 : For x∈(-π2,π2), if y(x)=∫cosecx+sinxcosecxsecx+tanxsin2x dx, and limx→(π2)-y(x)=0, then y(π4) is equal to [2024]

Q 6 : If ∫sin32x+cos32xsin3xcos3xsin(x-θ) dx=Acosθtanx-sinθ+Bcosθ-sinθcotx+C, where C is the integration constant, then AB is equal to [2024]

Q 7 : If ∫cosec5x dx=αcotxcosecx(cosec2x+32)+βloge|tanx2|+C, where α,β∈R and C is the constant of integration, then the value of 8(α+β) equals _______ . [2024]

Q 8 : If ∫1(x-1)4(x+3)65 dx=A(αx-1βx+3)B+C, where C is the constant of integration, then the value of α+β+20AB is ____ . [2024]

Q 9 : Let f(x)=∫x33–x2dx. If 5f(2)=–4, then f(1) is equal to [2025]

Q 10 : If ∫ex(x sin–1x1–x2+sin–1x((1–x2)3/2)+x1–x2)dx=g(x)+C, where C is the constant of integration, then g(12) equals: [2025]

Q 11 : Let I(x)=∫dx(x–11)1113(x+15)1513. If I(37)–I(24)=14(1b113–1c113), b, c∈N, then 3(b + c) is equal to [2025]

Q 12 : Let ∫x3 sin xdx=g(x)+C, where C is the constant of integration. If 8(g(π2)+g'(π2))=απ3+βπ2+γ, α, β, γ∈Z, then α+β–γ equals: [2025]

Q 13 : Let f:(0,∞)→R be a function which is differentiable at all points of its domain and satisfies the condition x2f'(x)=2xf(x)+3, with f(1) = 4. Then 2f(2) is equal to : [2025]

Q 14 : If f(x)=∫1x1/4(1+x1/4)dx, f(0) = – 6, then f(1) is equal to : [2025]

Q 15 : If ∫(1+x2+x)10(1+x2–x)9dx=1m((1+x2+x)n(n1+x2–x))+C where C is the constant of integration and m, n ∈ N, then m + n is equal to __________. [2025]

Q 16 : If ∫(1x+1x3)(3x–24+x–2623)dx=–α3(α+1)(3xβ+xγ)α+1α+C, x > 0, (α,β,γ∈Z), where c is the constant of integration, then α+β+γ is equal to __________. [2025]

Q 17 : If ∫2x2+5x+9x2+x+1dx=xx2+x+1+αx2+x+1+β loge |x+12+x2+x+1|+C, where C is the constant of integration, then α+2β is equal to __________. [2025]

Q 18 : Let I(x)=∫x2(xsec2x+tanx)(xtanx+1)2 dx. If I(0)=0, then I(π4) is equal to [2023]

Q 19 : Let I(x)=∫(x+1)x(1+xex)2dx, x>0. If limx→∞I(x)=0, then I(1) is equal to [2023]

Q 20 : For α,β,γ,δ∈ℕ, if ∫[(xe)2x+(ex)2x]logex dx=1α(xe)βx-1γ(ex)δx+C, where e=∑n=0∞1n!and C is constant of integration, then α+2β+3γ-4δ is equal to [2023]

Q 21 : Let f(x)=∫2x(x2+1)(x2+3)dx. If f(3)=12(loge5-loge6), then f(4) is equal to [2023]

Q 22 : Let I(x)=∫x+7xdx and I(9)=12+7loge7. If I(1)=α+7loge(1+22) then α4 is equal to _______ . [2023]

Q 23 : Let f(x)=∫dx(3+4x2)4-3x2,|x|<23. If f(0)=0 and f(1)=1αβtan-1(αβ), α,β>0, then α2+β2 is equal to _______ . [2023]

Q 24 : If ∫sec2x-1 dx=αloge|cos2x+β+cos2x(1+cos1βx)|+constant, then β-α is equal to _________ . [2023]

Q 25 : Let f(x)=∫(2-x2).ex(1+x)(1-x)3/2dx. If f(0)=0, then f(12) is equal to: [2026]

Q 26 : Let f(t)=∫(1-sin(loget)1-cos(loget))dt, t>1. If f(eπ/2)=-eπ/2 and f(eπ/4)=αeπ/4, then α equals [2026]

Q 27 : Let f(x)=∫dxx(23)+2x(12) be such that f(0)=-26+24loge(2). If f(1)=a+bloge(3), where a,b∈ℤ, then a+b is equal to: [2026]

Q 28 : If ∫(sinx)-112(cosx)-52dx=-p1q1(cotx)92-p2q2(cotx)52-p3q3(cotx)12+p4q4(cotx)-32+C where pi and qi are positive integers with gcd(pi,qi)=1 for i=1,2,3,4, and C is the constant of integration, then 15p1p2p3p4q1q2q3q4 is equal to ______ [2026]

Q 29 : Let f be a differentiable function satisfying f(x)=1-2x+∫0xe(x-t)f(t) dt, x∈ℝ and let g(x)=∫0x(f(t)+2)15(t-4)6(t+12)17dt, x∈ℝ. If p and q are respectively the points of local minima and local maxima of g, then the value of |p+q| is equal to ________ [2026]

Q 30 : Let f(x)=∫7x10+9x8(1+x2+2x9)2dx, x>0, limx→0f(x)=0 and f(1)=14. If A=[00114f'(1)1α241] and B=adj(adjA) be such that |B|=81, then α2 is equal to [2026]

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Q 1 :

$If \int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \frac{1}{12} \tan^{- 1} (3 \tan x) + constant,$ $then the maximum value of a \sin x + b \cos x is:$ [2024]

Q 2 :

Let $I (x) = \int \frac{6}{\sin^{2} x {(1 - \cot x)}^{2}} d x .$ If $I (0) = 3,$ then $I (\frac{π}{12})$ is equal to [2024]

Q 3 :

Let $\int \frac{2 - \tan x}{3 + \tan x} d x = \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C,$ where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to: [2024]

Q 4 :

The integral $\int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$ is equal to: [2024]

Q 5 :

$For x \in (- \frac{π}{2}, \frac{π}{2}), if y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x,$ $and \lim_{x \to {(\frac{π}{2})}^{-}} y (x) = 0, then y (\frac{π}{4}) is equal to$ [2024]

Q 6 :

If $\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^{3} x \cos^{3} x \sin (x - θ)}} d x = A \sqrt{\cos θ \tan x - \sin θ} + B \sqrt{\cos θ - \sin θ \cot x} + C,$ where C is the integration constant, then AB is equal to [2024]

Q 7 :

If $\int c o s e c^{5} x d x = α \cot x c o s e c x (c o s e c^{2} x + \frac{3}{2}) + β \log_{e} | \tan \frac{x}{2} | + C,$ where $α, β \in R$ and $C$ is the constant of integration, then the value of $8 (α + β)$ equals _______ . [2024]

Q 8 :

$If \int \frac{1}{\sqrt[5]{{(x - 1)}^{4} {(x + 3)}^{6}}} d x = A {(\frac{α x - 1}{β x + 3})}^{B} + C, where C is the constant of integration,$ $then the value of α + β + 20 A B is ____ .$ [2024]

Q 9 :

Let $f (x) = \int x^{3} \sqrt{3 - x^{2}} d x$ . If $5 f (\sqrt{2}) = - 4$ , then f(1) is equal to [2025]

Q 10 :

If $\int e^{x} (\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{\sin^{- 1} x}{({(1 - x^{2})}^{3 / 2})} + \frac{x}{1 - x^{2}}) d x = g (x) + C$ , where C is the constant of integration, then $g (\frac{1}{2})$ equals: [2025]

Q 11 :

Let $I (x)$ $= \int \frac{d x}{{(x - 11)}^{\frac{11}{13}} {(x + 15)}^{\frac{15}{13}}}$ . If $I (37) - I (24) = \frac{1}{4} (\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}})$ , $b, c \in N$ , then 3(b + c) is equal to [2025]

Q 12 :

Let $\int x^{3} \sin x d x = g (x) + C$ , where C is the constant of integration. If $8 (g (\frac{π}{2}) + g' (\frac{π}{2})) = α π^{3} + {βπ}^{2} + γ, α, β, γ \in Z$ , then $α + β - γ$ equals: [2025]

Q 13 :

Let $f : (0, \infty) \to R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f' (x) = 2 x f (x) + 3$ , with f(1) = 4. Then 2f(2) is equal to : [2025]

Q 14 :

If $f (x) = \int \frac{1}{x^{1 / 4} (1 + x^{1 / 4})} d x$ , f(0) = – 6, then f(1) is equal to : [2025]

Q 15 :

If $\int \frac{{(\sqrt{1 + x^{2}} + x)}^{10}}{{(\sqrt{1 + x^{2}} - x)}^{9}} d x = \frac{1}{m} ({(\sqrt{1 + x^{2}} + x)}^{n} (n \sqrt{1 + x^{2}} - x)) + C$ where C is the constant of integration and m, n $\in$ N, then m + n is equal to __________. [2025]

Q 16 :

If $\int (\frac{1}{x} + \frac{1}{x^{3}}) (\sqrt[23]{3 x^{- 24} + x^{- 26}}) d x = - \frac{α}{3 (α + 1)} {(3 x^{β} + x^{γ})}^{\frac{α + 1}{α}} + C$ , x > 0, $(α, β, γ \in Z)$ , where c is the constant of integration, then $α + β + γ$ is equal to __________. [2025]

Q 17 :

If $\int \frac{2 x^{2} + 5 x + 9}{\sqrt{x^{2} + x + 1}} d x = x \sqrt{x^{2} + x + 1} + α \sqrt{x^{2} + x + 1} + β \log_{e}$ $| x + \frac{1}{2} + \sqrt{x^{2} + x + 1} |$ $+ C$ , where C is the constant of integration, then $α + 2 β$ is equal to __________. [2025]

Q 18 :

$Let I (x) = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x . If I (0) = 0, then I (\frac{π}{4}) is equal to$ [2023]

Q 19 :

$Let I (x) = \int \frac{(x + 1)}{x {(1 + x e^{x})}^{2}} d x, x > 0 . If \lim_{x \to \infty} I (x) = 0, then I (1) is equal to$ [2023]

Q 21 :

$Let f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x .$ $If f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6), then f (4) is equal to$ [2023]

Q 22 :

Let $I (x) = \int \sqrt{\frac{x + 7}{x}} d x$ and $I (9) = 12 + 7 \log_{e} 7$ . If $I (1) = α + 7 \log_{e} (1 + 2 \sqrt{2})$ then $α^{4}$ is equal to _______ . [2023]

Q 23 :

Let $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}},$ $| x |$ $< \frac{2}{\sqrt{3}}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , $α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______ . [2023]

Q 24 :

If $\int \sqrt{\sec 2 x - 1} d x = α \log_{e}$ $| \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} |$ $+$ constant, then $β - α$ is equal to _________ . [2023]

Q 25 :

Let $f (x)$ $= \int \frac{(2 - x^{2}) . e^{x}}{(\sqrt{1 + x}) {(1 - x)}^{3 / 2}} d x .$ If $f (0) = 0$ , then $f (\frac{1}{2})$ is equal to: [2026]

Q 26 :

Let $f (t)$ $= \int (\frac{1 - \sin (\log_{e} t)}{1 - \cos (\log_{e} t)}) d t, t > 1 .$

If $f (e^{π / 2}) = - e^{π / 2} and f (e^{π / 4}) = α e^{π / 4},$ then $α$ equals [2026]

Q 27 :

Let $f (x)$ $= \int \frac{d x}{x^{(\frac{2}{3})} + 2 x^{(\frac{1}{2})}}$ be such that $f (0) = - 26 + 24 \log_{e} (2) .$ If $f (1) = a + b \log_{e} (3),$ where $a, b \in ℤ$ , then a+b is equal to: [2026]