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

Q 11 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

(2)

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

$f (x) = e^{x} + A + B x e^{x}$

$A = \int_{0}^{1} y f (y) d y = \int_{0}^{1} y (A + e^{y} + B y e^{y}) d y$

$A = \frac{A}{2} + 0 - (- 1) + B (e - 1)$

$\frac{A}{2} + B (1 - e) = 1$

$B = \int_{0}^{1} f (y) d y$

$B = \int_{0}^{1} (e^{y} + A + B y e^{y}) d y$

$B = (e - 1) + A + B (0 - (- 1))$

$B = e - 1 + A + B \Rightarrow A = 1 - e$

$f (x) = e^{x} + A + B x e^{x}$

$f (0) = 1 + A = 1 - e + 1 = 2 - e$

$e + f (0) = 2$

Q 12 :

Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

31
30
29
28

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(1)

$Let 4 x + 6 = \frac{1}{t} \Rightarrow x = \frac{\frac{1}{t} - 6}{4}$

$4 d x = - \frac{d t}{t^{2}},$ ${\frac{x + 1 = \frac{1}{t} - 2}{4}$

$\int \frac{3 d x}{(4 x + 6) \sqrt{4 {(x + 1)}^{2} - 1}}$

$= \int \frac{3 (- d t)}{4 t^{2} \cdot \frac{1}{t} \sqrt{4 {(\frac{1 / t - 2}{4})}^{2} - 1}}$

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

$= - \frac{3}{4} \int \frac{d t (2 t)}{t \sqrt{1 - 4} t}$

$= - \frac{3}{2} \int \frac{d t}{\sqrt{1 - 4 t}}$ $= - \frac{3}{2} (\frac{\sqrt{1 - 4 t}}{\frac{1}{2} \times - 4}) + C$

$= \frac{3}{4} \sqrt{1 - 4 t} + C$ $∵$ $t = \frac{1}{4 x + 6}$

$= \frac{3}{4} \sqrt{1 - 4 (\frac{1}{4 x + 6})} + C$

$= \frac{3}{4} \sqrt{\frac{4 x + 6 - 4}{4 x + 6}} + C$

$I (x) = \frac{3}{4} \sqrt{\frac{4 x + 2}{4 x + 6}} + C$

$I (0) = \frac{3}{4} \sqrt{\frac{2}{6}} + C = \frac{\sqrt{3}}{4} + C$

$\Rightarrow C = 20$

$Hence I (x) = \frac{3}{4} \sqrt{\frac{4 x + 2}{4 x + 6}} + 20$

$I (\frac{1}{2}) = \frac{3}{4} \sqrt{\frac{4}{8}} + 20$

$= \frac{3}{4 \sqrt{2}} + 20 = \frac{3 \sqrt{2}}{8} + 20$

$a + b + c = 3 + 8 + 20 = 31$

Q 13 :

If $\int ((\frac{1 - 5 \cos^{2} x}{\sin^{5} x \cos^{2} x})) d x = f (x) + C,$ where C is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to [2026]

$\frac{4}{\sqrt{3}} (8 - \sqrt{6})$
$\frac{2}{\sqrt{3}} (4 + \sqrt{6})$
$\frac{1}{\sqrt{3}} (26 + \sqrt{3})$
$\frac{1}{\sqrt{3}} (26 - \sqrt{3})$

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(1)

$\int \frac{d x}{\sin^{5} x \cos^{2} x} - 5 \int \frac{d x}{\sin^{5} x}$

$= \int \frac{\sec^{2} x d x}{\sin^{5} x} - 5 \int \frac{d x}{\sin^{5} x}$

$By IBP$

$= \frac{\tan x}{\sin^{5} x} - \int \frac{5}{\sin^{6} x} \cos x \tan x d x - 5 \int \frac{d x}{\sin^{5} x}$

$= \frac{\tan x}{\sin^{5} x} + C$

$f (x) = \frac{\tan x}{\sin^{5} x}$

$f (\frac{π}{6}) - f (\frac{π}{4}) = \frac{2^{5}}{\sqrt{3}} - {(\sqrt{2})}^{5} = 4 \sqrt{2} - \frac{32}{\sqrt{3}}$

$= \frac{32}{\sqrt{3}} - 4 \sqrt{2}$

$= \frac{4}{\sqrt{3}} (8 - \sqrt{6})$

Q 14 :

$\int \frac{6 x^{10} + 4}{x^{3} \sqrt{x^{10} - 3 x^{4} - 1}} d x, x > 0 =$ _____________ (c is constant of integration)

$\frac{\sqrt{x^{10} - 3 x^{4} - 1}}{x^{2}} + c$
$\frac{2 \sqrt{x^{10} - 3 x^{4} - 1}}{x^{2}} + c$
$2 \sqrt{x^{10} - 3 x^{4} - 1} + c$
$\sqrt{x^{10} - 3 x^{4}} + C$

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(2)

$\int \frac{6 x^{10} + 4}{x^{5} \sqrt{x^{6} - 3 - x^{- 4}}} d x = \int \frac{6 x^{5} + 4 x^{- 5}}{\sqrt{x^{6} - 3 - x^{- 4}}} d x$

$x^{6} - 3 - x^{- 4} = t (6 x^{5} + 4 x^{- 5}) d x = d t$

Q 15 :

If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

(2)

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

$(x^{2} + 1) e^{x} = t$

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

$I = \frac{1}{2} ({(x^{2} + 1) e^{x})}^{2} + c$

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

$2 A + f (0) = 1 + 1 = 2$

Q 16 :

$\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to:

$2 (\sin x + x \cos θ) + C$
$2 (\sin x - x \cos θ) + C$
$2 (\sin x + 2 x \cos θ) + C$
$2 (\sin x - 2 x \cos θ) + C$

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(1)

Ans. $2 (s i n x + x c o s θ) + C$

Explanation:

Let $I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$

$= \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} d x$

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

$= 2 \int \frac{(\cos x - \cos θ) (\cos x + \cos θ)}{\cos x - \cos θ} d x$

$= 2 \int (\cos x + \cos θ) d x$

$= 2 (\sin x + x \cos θ) + C$

Q 17 :

$\int \frac{d x}{\sin (x - a) \sin (x - b)}$ is equal to:

$\sin (b - a) \log$ $| \frac{\sin (x - b)}{\sin (x - a)} |$ $+ C$
$c o s e c (b - a) \log$ $| \frac{\sin (x - a)}{\sin (x - b)} |$ $+ C$
$c o s e c (b - a) \log$ $| \frac{\sin (x - b)}{\sin (x - a)} |$ $+ C$
$\sin (b - a) \log$ $| \frac{\sin (x - a)}{\sin (x - b)} |$ $+ C$

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(3)

Ans. $c o s e c (b - a) \log$ $| \frac{\sin (x - b)}{\sin (x - a)} |$ $+ C$

Explanation:

$Let I = \int \frac{d x}{\sin (x - a) \sin (x - b)}$

$= \frac{1}{\sin (b - a)} \int \frac{\sin (b - a)}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \int \frac{\sin (x - a - x + b)}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \int \frac{\sin {(x - a) - (x - b)}}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \times \int \frac{\sin (x - a) \cos (x - b) - \cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)} d x$

$= \frac{1}{\sin (b - a)} \int [\cot (x - b) - \cot (x - a)] d x$

$= \frac{1}{\sin (b - a)} [\log | \sin (x - b) | - \log | \sin (x - a) |] + C$

$= cosec (b - a) \log$ $| \frac{\sin (x - b)}{\sin (x - a)} |$ $+ C$

Q 18 :

$\int \frac{x^{3}}{x + 1} d x$ is equal to:

$x + \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log$ $| 1 - x |$ $+ C$
$x + \frac{x^{2}}{2} - \frac{x^{3}}{3} - \log$ $| 1 - x |$ $+ C$
$x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \log$ $| 1 + x |$ $+ C$
$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log$ $| 1 + x |$ $+ C$

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(4)

Ans. $x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log$ $| 1 + x |$ $+ C$

Explanation:

$Let$ $I = \int \frac{x^{3}}{x + 1} d x$

$= \int \frac{x^{3} + 1 - 1}{x + 1} d x$

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

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

$= \frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \log$ $| x + 1 |$ $+ C$

Q 19 :

$\int 7 e^{7 x + 5} d x =$

$7 (7 e^{7 x + 5}) + C$
$\frac{7 e^{7 + 5}}{5} + C$
$e^{7 x + 5} + C$
$\frac{7 e^{7 + 5}}{7 x + 5} + C$

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(3)

Ans. $e^{7 x + 5} + C$

Explanation:

$\int 7 e^{7 x + 5} d x = 7 \cdot \frac{e^{7 x + 5}}{7} + C = e^{7 x + 5} + C$

Q 20 :

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

$\log x + C$
$(\log x) (\tan^{- 1} x) + C$
$2 \log x + \tan^{- 1} x + C$
$\log x + 2 \tan^{- 1} x + C$

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(4)

Ans. $l o g x + 2 t a n^{- 1} x + C$

Explanation

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

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

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

$= \log x + 2 \tan^{- 1} x + C$

Q 21 :

If $f' (x) = \frac{1}{x} + x$ and $f (1) = \frac{5}{2}$ , then $f (x) =$

$\log x + \frac{x^{2}}{2} + 2$
$\log x + \frac{x^{2}}{2} + 1$
$\log x - \frac{x^{2}}{2} + 2$
$\log x - \frac{x^{2}}{2} + 1$

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(1)

Ans. $\log x + \frac{x^{2}}{2} + 2$

Explanation

$f (x) = \int f' (x) d x = \int (\frac{1}{x} + x) d x$

$= \log x + \frac{x^{2}}{2} + C$

$f (1) = \log 1 + \frac{1^{2}}{2} + C$

$\Rightarrow$ $\frac{5}{2} = 0 + \frac{1}{2} + C$

$\Rightarrow$ $C = 2$

$∴$ $f (x) = \log x + \frac{x^{2}}{2} + 2$

Q 22 :

$\int 2^{x} 3^{x + 1} 4^{x + 2} d x =$

$\frac{{(48)}^{x}}{\log 48} + C$
$\frac{2^{x} 3^{x + 1} 4^{x + 2}}{\log 2 + \log 4 + \log 3} + C$
$\frac{{(24)}^{x + 2}}{\log 24} + C$
$\frac{2^{x + 1} 3^{x + 2} 4^{x + 3}}{\log 2 + \log 4 + \log 3} + C$

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Ans. $\frac{2^{x} 3^{x + 1} 4^{x + 2}}{\log 2 + \log 4 + \log 3} + C$

Explanation

$\int 2^{x} 3^{x + 1} 4^{x + 2} d x = 16 \times 3 \int 2^{x} 3^{x} 4^{x} d x$

$= 48 \int {(24)}^{x} d x = \frac{48 {(24)}^{x}}{\log 24} + C$

$= \frac{2^{x} 3^{x + 1} 4^{x + 2}}{\log 2 + \log 4 + \log 3} + C$

Q 23 :

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

$\sin x + C$
$\cos x + C$
$\tan x + C$
$0$

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(3)

Ans. $\tan x + C$

Explanation:

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

$= \int \frac{d x}{\cos^{2} x} = \int \sec^{2} x d x$

$= \tan x + C$

Q 24 :

$\int x^{51} (\tan^{- 1} x + \cot^{- 1} x) d x =$

$\frac{x^{52}}{52} (\tan^{- 1} x + \cot^{- 1} x) + C$
$\frac{x^{52}}{52} (\tan^{- 1} x - \cot^{- 1} x) + C$
$\frac{π x^{52}}{104} + \frac{π}{2} + C$
$\frac{x^{52}}{52} + \frac{π}{2} + C$

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Ans. $\frac{x^{52}}{52} (\tan^{- 1} x + \cot^{- 1} x) + C$

Explanation

$\int x^{51} (\tan^{- 1} x + \cot^{- 1} x) d x$

$= \int x^{51} \cdot \frac{π}{2} d x$ $[∵ \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}]$

$= \frac{π x^{52}}{104} + C$ $= \frac{x^{52}}{52} (\tan^{- 1} x + \cot^{- 1} x) + C$

Q 25 :

The value of $\int \frac{1}{1 + \cos 8 x} d x$ is:

$\frac{\tan 2 x}{8} + C$
$\frac{\tan 8 x}{8} + C$
$\frac{\tan 4 x}{4} + C$
$\frac{\tan 4 x}{8} + C$

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Ans. $\frac{\tan 4 x}{8} + C$

Explanation:

$\int \frac{1}{1 + \cos 8 x} d x = \int \frac{d x}{1 + \frac{(1 - \tan^{2} 4 x)}{(1 + \tan^{2} 4 x)}}$

$= \int \frac{1 + \tan^{2} 4 x}{1 + \tan^{2} 4 x + 1 - \tan^{2} 4 x} d x$

$= \int \frac{1 + \tan^{2} 4 x}{2} d x$

$= \frac{1}{2} \int \sec^{2} 4 x d x$

$= \frac{\tan 4 x}{8} + C$

Q 26 :

By the inspection method, evaluate:

$\int (a u^{2} + b u + c) d u$

$a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) + c u + C$
$- a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) - c u + C$
$a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) - c u + C$
$a (\frac{u^{3}}{3}) - b (\frac{u^{2}}{2}) - c u + C$

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Ans. $a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) + c u + C$

Explanation:

The given integral is $\int (a u^{2} + b u + c) d u$

$\int (a u^{2} + b u + c) d u = \int a u^{2} d u + \int b u d u + \int c d u$

$= a \int u^{2} d u + b \int u d u + c \int d u$

$= a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) + c u + C$

Q 27 :

Find the integral of $\sqrt{u} + \frac{1}{\sqrt{u}} .$

$\frac{2}{3} u^{3 / 2} + 2 u^{1 / 2} + C$
$\frac{2}{3} u^{3 / 2} - 2 u^{1 / 2} + C$
$- \frac{2}{3} u^{3 / 2} - 2 u^{1 / 2} + C$
$- \frac{2}{3} u^{3 / 2} + 2 u^{1 / 2} + C$

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Ans. $\frac{2}{3} u^{3 / 2} + 2 u^{1 / 2} + C$

Explanation:

$\int (\sqrt{u} + \frac{1}{\sqrt{u}}) d u = \int (\sqrt{u}) d u + \int (\frac{1}{\sqrt{u}}) d u$

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

$= \frac{u^{3 / 2}}{\frac{3}{2}} + \frac{u^{1 / 2}}{\frac{1}{2}} + C$

$= \frac{2}{3} u^{3 / 2} + 2 u^{1 / 2} + C$

Q 28 :

$\int 4^{x} 3^{x} d x$ equals:

$\frac{12^{x}}{\log 12} + C$
$\frac{4^{x}}{\log 4} + C$
$(\frac{4^{x} \cdot 3^{x}}{\log 4 \cdot \log 3}) + C$
$\frac{3^{x}}{\log 3} + C$

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(1)

Ans. $\frac{12^{x}}{\log 12} + C$

Explanation:

Let $I = \int 4^{x} 3^{x} d x$

$\Rightarrow$ $I = \int 12^{x} d x$

$\Rightarrow$ $I = \frac{12^{x}}{\log 12} + C$

Q 29 :

Find the integral of $\sin u \sin 2 u \sin 3 u$ .

$\frac{1}{4} [\cos 6 u - \cos 4 u - \cos 2 u] + C$
$\frac{1}{4} [\cos 4 u - \cos 6 u - \cos 2 u] + C$
$\frac{1}{8} [\frac{\cos 6 u}{3} - \frac{\cos 4 u}{2} - \cos 2 u] + C$
$\frac{1}{8} [\frac{\cos 6 u}{3} - \frac{\cos 4 u}{2} + \cos 2 u] + C$

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Ans. $\frac{1}{8} [\frac{\cos 6 u}{3} - \frac{\cos 4 u}{2} - \cos 2 u] + C$

Explanation:

The given integral is,

$\int \sin u \sin 2 u \sin 3 u d u .$

We know that,

$\sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)]$

Using the above formula for $\sin 2 u \sin 3 u$ in the given integral, we get

$\int \sin u \sin 2 u \sin 3 u d u$

$= \int \sin u \cdot \frac{1}{2} [\cos (2 u - 3 u) - \cos (2 u + 3 u)] d u$

$= \frac{1}{2} \int [\sin u \cos (- u) - \sin u \cos 5 u] d u$

$= \frac{1}{2} \int (\sin u \cos u - \sin u \cos 5 u) d u$

We know that,

$2 \sin A \cos A = \sin 2 A$

and $\sin A \cos B = \frac{1}{2} [\sin (A + B) + \sin (A - B)]$

Using the above two formulas in the integral, we get

$\frac{1}{2} \int (\sin u \cos u - \sin u \cos 5 u) d u$

$= \frac{1}{2} \int \frac{2 \sin u \cos u}{2} d u - \frac{1}{2} \int \frac{1}{2} [\sin (u + 5 u) + \sin (u - 5 u)] d u$

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

$= \frac{1}{4} [\int \sin 2 u d u - {\int \sin 6 u d u - \int \sin 4 u d u}]$

$= \frac{1}{4} [\frac{- \cos 2 u}{2} - (\frac{- \cos 6 u}{6} + \frac{\cos 4 u}{4})] + C$

$= \frac{1}{4} \times \frac{1}{2} [- \cos 2 u + \frac{\cos 6 u}{3} - \frac{\cos 4 u}{2}] + C$

$= \frac{1}{8} [\frac{\cos 6 u}{3} - \frac{\cos 4 u}{2} - \cos 2 u] + C$

Q 30 :

$\int \tan^{- 1} \sqrt{x} d x$ is equal to:

$(x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$
$x \tan^{- 1} x - \sqrt{x} + C$
$\sqrt{x} - x \tan^{- 1} \sqrt{x} + C$
$\sqrt{x} - (x + 1) \tan^{- 1} \sqrt{x} + C$

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Ans. $(x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$

Explanation:

Let $I = \int 1 \cdot \tan^{- 1} \sqrt{x} d x$

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

$= x \tan^{- 1} \sqrt{x} - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} d x$

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

$∴$ $I = x \tan^{- 1} \sqrt{x} - \frac{1}{2} \int \frac{t \cdot 2 t}{1 + t^{2}} d t$

$= x \tan^{- 1} \sqrt{x} - \int \frac{t^{2}}{1 + t^{2}} d t$

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

$= x \tan^{- 1} \sqrt{x} - t + \tan^{- 1} t + C$

$= x \tan^{- 1} \sqrt{x} - \sqrt{x} + \tan^{- 1} \sqrt{x} + C$ $(\begin{matrix} ∵ x = t^{2} \\ ∴ t = \sqrt{x} \end{matrix})$

$= (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$

Q 31 :

$\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$ is equal to:

$\frac{1}{5 x} {(4 + \frac{1}{x^{2}})}^{- 5} + C$
$\frac{1}{5} {(4 + \frac{1}{x^{2}})}^{- 5} + C$
$\frac{1}{10 x} {(1 + 4)}^{- 5} + C$
$\frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

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Ans. $\frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

Explanation:

Let $I = \int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$

$= \int \frac{x^{9}}{x^{12} {(4 + \frac{1}{x^{2}})}^{6}} d x$

$= \int \frac{d x}{x^{3} {(4 + \frac{1}{x^{2}})}^{6}}$

Put $4 + \frac{1}{x^{2}} = t$

$\Rightarrow$ $\frac{- 2}{x^{3}} d x = d t$

$\Rightarrow$ $\frac{1}{x^{3}} d x = - \frac{1}{2} d t$

$∴$ $I = - \frac{1}{2} \int \frac{d t}{t^{6}} = \frac{- 1}{2} \int t^{- 6} d t$

$= - \frac{1}{2} [\frac{t^{- 6 + 1}}{- 6 + 1}] + C$

$= \frac{1}{10} [\frac{1}{t^{5}}] + C$ $= \frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

Q 32 :

If $\int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = a {(1 + x^{2})}^{3 / 2} + b \sqrt{1 + x^{2}} + C,$ then:

$a = \frac{1}{3}, b = 1$
$a = \frac{- 1}{3}, b = 1$
$a = \frac{- 1}{3}, b = - 1$
$a = \frac{1}{3}, b = - 1$

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Ans. $a = \frac{1}{3}, b = - 1$

Explanation:

Let $I = \int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x$

$= a {(1 + x^{2})}^{3 / 2} + b \sqrt{1 + x^{2}} + C$

$∵$ $I = \int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x$

$= \int \frac{x^{2} \cdot x}{\sqrt{1 + x^{2}}} d x$

Put $1 + x^{2} = t^{2}$

$\Rightarrow$ $2 x d x = 2 t d t$

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

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

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

By comparison, $a = \frac{1}{3}$ and $b = - 1$

Q 33 :

$\int \frac{1}{\log a} (a^{x} \cos a^{x}) d x =$

$\sin a^{x} + C$
$a^{x} \sin a^{x} + C$
$\frac{1}{{(\log a)}^{2}} \sin a^{x} + C$
$\log (\sin a^{x}) + C$

1

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(3)

Ans. $\frac{1}{{(\log a)}^{2}} \sin a^{x} + C$

Explanation:

Put $a^{x} = t$

Then, $a^{x} \log a d x = d t$

$\Rightarrow$ $a^{x} d x = \frac{d t}{\log a}$

Therefore, $\int \frac{1}{\log a} (a^{x} \cos a^{x}) d x = \int \frac{1}{{(\log a)}^{2}} \cos t d t$

$= \frac{1}{{(\log a)}^{2}} \sin t + C$

$= \frac{1}{{(\log a)}^{2}} \sin a^{x} + C$

Q 34 :

Find the integral of $\frac{5 x^{4}}{\sqrt{x^{5} + 9}}$ :

$\sqrt{x^{5} + 9} + C$
$2 \sqrt{x^{5} - 9} + C$
$2 (x^{5} + 9) + C$
$2 \sqrt{x^{5} + 9} + C$

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(4)

Ans. $2 \sqrt{x^{5} + 9} + C$

Explanation:

Let $x^{5} + 9 = t$

Differentiating with respect to $x$ , we get

$5 x^{4} d x = d t$

$\int \frac{5 x^{4}}{\sqrt{x^{5} + 9}} d x = \int \frac{d t}{\sqrt{t}}$

$= \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C$

$= 2 \sqrt{t} + C$

Replacing $t$ with $x^{5} + 9$ , we get

$\int \frac{5 x^{4}}{\sqrt{x^{5} + 9}} d x = 2 \sqrt{x^{5} + 9} + C$

Q 35 :

$\int \frac{\sec x d x}{\sqrt{\cos 2 x}} =$

$\sin^{- 1} (\tan x)$
$\tan x$
$\cos^{- 1} (\tan x)$
$\frac{\sin x}{\sqrt{\cos x}}$

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2

3

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(1)

Ans. $\sin^{- 1} (\tan x)$

Explanation:

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

$= \int \frac{\sec^{2} x d x}{\sqrt{1 - \tan^{2} x}}$

$(Multiplying numerator and denominator by \sec x)$

Now put $\tan x = t$

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

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

$= \sin^{- 1} (\tan x)$

Q 36 :

If $\int x^{6} \sin (5 x^{7}) d x = \frac{k}{5} \cos (5 x^{7}) + C, x \neq 0,$ then:

$k = 7$
$k = - 7$
$k = \frac{1}{7}$
$k = - \frac{1}{7}$

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(4)

Ans. $k = - \frac{1}{7}$

Explanation:

Put $5 x^{7} = t$

$\Rightarrow$ $35 x^{6} d x = d t$

$\Rightarrow$ $x^{6} d x = \frac{d t}{35}$

Therefore, $\int x^{6} \sin (5 x^{7}) d x = \int \sin t \cdot \frac{d t}{35}$

$= - \frac{\cos t}{35} + C$

$= - \frac{\cos (5 x^{7})}{35} + C$

Comparing with

$\int x^{6} \sin (5 x^{7}) d x = \frac{k}{5} \cos (5 x^{7}) + C,$

we get $\frac{k}{5} = - \frac{1}{35}$

$∴$ $k = - \frac{1}{7}$

Q 37 :

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

$- \frac{1}{2 (e^{2 x} + 1)} + C$
$\frac{1}{2 (e^{2 x} + 1)} + C$
$- \frac{1}{e^{2 x} + 1} + C$
None of these

1

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3

4

(1)

Ans. $- \frac{1}{2 (e^{2 x} + 1)} + C$

Explanation:

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

Put $e^{2 x} + 1 = t$ $\Rightarrow 2 e^{2 x} d x = d t,$

then it reduces to

$\frac{1}{2} \int \frac{1}{t^{2}} d t = - \frac{1}{2} \cdot \frac{1}{t} + C$

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

Q 38 :

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

$\log a \cdot a^{x + \tan^{- 1} x} + C$
$\frac{(x + \tan^{- 1} x)}{\log a} + C$
$\frac{a^{x + \tan^{- 1} x}}{\log a} + C$
$\log a \cdot (x + \tan^{- 1} x) + C$

1

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(3)

Ans. $\frac{a^{x + \tan^{- 1} x}}{\log a} + C$

Explanation :

$Put$ $x + \tan^{- 1} x = t$

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

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

$\int [\frac{(x^{2} + 2) a^{(x + \tan^{- 1} x)}}{x^{2} + 1}] d x = \int a^{t} d t$

$= \frac{a^{t}}{\log a} + C$

$= \frac{a^{x + \tan^{- 1} x}}{\log a} + C$

Q 39 :

$\int \frac{1}{1 + \cos^{2} x} d x =$

$\frac{1}{\sqrt{2}} \tan^{- 1} (\tan x) + C$
$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{2} \tan x) + C$
$\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} \tan x) + C$
None of the above

1

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(3)

Ans. $\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} \tan x) + C$

Explanation :

$\int \frac{d x}{1 + \cos^{2} x} = \int \frac{\sec^{2} x}{\sec^{2} x + 1} d x = \int \frac{\sec^{2} x}{\tan^{2} x + 2} d x$

$= \int \frac{d t}{t^{2} + 2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t}{\sqrt{2}}) + C$ ${Putting \tan x = t, \sec^{2} x d x = d t}$

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

Q 40 :

Evaluate: $\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u$

$\log (10^{u} + u^{10}) + C$
$(10^{u} - u^{10}) + C$
$10^{u} + u^{10} + C$
$- 10^{u} + u^{10} + C$

1

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(1)

Ans. $\log (10^{u} + u^{10}) + C$

Explanation

$The given integral is$ $\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u$

$Let$ $u^{10} + 10^{u} = v$

$Differentiating both sides, we get$

$\frac{d}{d u} (u^{10} + 10^{u}) = \frac{d v}{d u}$

$10 u^{9} + 10^{u} \log_{e} 10 = \frac{d v}{d u}$

$(10 u^{9} + 10^{u} \log_{e} 10) d u = d v$

$Substitute v for u^{10} + 10^{u} and d v for (10 u^{9} + 10^{u} \log_{e} 10) d u in the given integral.$

$\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u = \int \frac{d v}{v}$

$= \int \frac{1}{v} d v$

$= \log v + C$

$= \log (u^{10} + 10^{u}) + C$ $(∵ v = u^{10} + 10^{u})$

Topic Question Set

Q 11 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

Q 12 :

Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

Q 13 :

If $\int ((\frac{1 - 5 \cos^{2} x}{\sin^{5} x \cos^{2} x})) d x = f (x) + C,$ where C is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to [2026]

Q 14 :

$\int \frac{6 x^{10} + 4}{x^{3} \sqrt{x^{10} - 3 x^{4} - 1}} d x, x > 0 =$ _____________ (c is constant of integration)

Q 15 :

If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

Q 16 :

$\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to:

Q 17 :

$\int \frac{d x}{\sin (x - a) \sin (x - b)}$ is equal to:

Q 18 :

$\int \frac{x^{3}}{x + 1} d x$ is equal to:

Q 19 :

$\int 7 e^{7 x + 5} d x =$

Q 20 :

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

Q 21 :

If $f' (x) = \frac{1}{x} + x$ and $f (1) = \frac{5}{2}$ , then $f (x) =$

Q 22 :

$\int 2^{x} 3^{x + 1} 4^{x + 2} d x =$

Q 23 :

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

Q 24 :

$\int x^{51} (\tan^{- 1} x + \cot^{- 1} x) d x =$

Q 25 :

The value of $\int \frac{1}{1 + \cos 8 x} d x$ is:

Q 26 :

By the inspection method, evaluate:

$\int (a u^{2} + b u + c) d u$

Q 27 :

Find the integral of $\sqrt{u} + \frac{1}{\sqrt{u}} .$

Q 28 :

$\int 4^{x} 3^{x} d x$ equals:

Q 29 :

Find the integral of $\sin u \sin 2 u \sin 3 u$ .

Q 30 :

$\int \tan^{- 1} \sqrt{x} d x$ is equal to:

Q 31 :

$\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$ is equal to:

Q 32 :

If $\int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = a {(1 + x^{2})}^{3 / 2} + b \sqrt{1 + x^{2}} + C,$ then:

Q 33 :

$\int \frac{1}{\log a} (a^{x} \cos a^{x}) d x =$

Q 34 :

Find the integral of $\frac{5 x^{4}}{\sqrt{x^{5} + 9}}$ :

Q 35 :

$\int \frac{\sec x d x}{\sqrt{\cos 2 x}} =$

Q 36 :

If $\int x^{6} \sin (5 x^{7}) d x = \frac{k}{5} \cos (5 x^{7}) + C, x \neq 0,$ then:

Q 37 :

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

Q 38 :

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

Q 39 :

$\int \frac{1}{1 + \cos^{2} x} d x =$

Q 40 :

Evaluate: $\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u$

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 11 : If f(x) satisfies the relation f(x)=ex+∫01(y+xex) f(y) dy, then e+f(0) is equal to _______ . [2026]

Q 12 : Let I(x)=∫3dx(4x+6)(4x2+8x+3) and I(0)=34+20. If I (12)=a2b+c, where a,b,c∈ℕ, gcd(a,b)=1,then a+b+c is equal to. [2026]

Q 13 : If ∫(1-5cos2xsin5x cos2x)dx=f(x)+C, where C is the constant of integration, then f(π6)-f(π4) is equal to [2026]

Q 14 : ∫6x10+4x3x10-3x4-1 dx, x>0= _____________ (c is constant of integration)

Q 15 : If I=∫(x2+1)((x+1)ex)2dx=A(f(x))2+C, where C is constant of integration and f(-1)=2e, then 2A+f(0) is

Q 16 : ∫cos2x-cos2θcosx-cosθdx is equal to:

Q 17 : ∫dxsin(x-a)sin(x-b) is equal to:

Q 18 : ∫x3x+1dx is equal to:

Q 19 : ∫7e7x+5 dx=

Q 20 : ∫(1+x)2x(1+x2) dx

Q 21 : If f'(x)=1x+x and f(1)=52, then f(x)=

Q 22 : ∫2x 3x+1 4x+2 dx=

Q 23 : ∫dxcos2x+sin2x=

Q 24 : ∫x51(tan-1x+cot-1x)dx=

Q 25 : The value of ∫11+cos8xdx is:

Q 26 : By the inspection method, evaluate: ∫(au2+bu+c)du

Q 27 : Find the integral of u+1u.

Q 28 : ∫4x3x dx equals:

Q 29 : Find the integral of sinu sin2u sin3u.

Q 30 : ∫tan-1x dx is equal to:

Q 31 : ∫x9(4x2+1)6 dx is equal to:

Q 32 : If ∫x31+x2 dx=a(1+x2)3/2+b1+x2+C, then:

Q 33 : ∫1loga(ax cos ax) dx=

Q 34 : Find the integral of 5x4x5+9 :

Q 35 : ∫secx dxcos2x=

Q 36 : If ∫x6sin(5x7) dx=k5cos(5x7)+C, x≠0, then:

Q 37 : ∫1(ex+e-x)2 dx=

Q 38 : ∫((x2+2)a(x+tan-1x)x2+1) dx=

Q 39 : ∫11+cos2x dx=

Q 40 : Evaluate: ∫10u9+10uloge10u10+10u du

Want Us to Email you About Special Offers & Updates?

Q 11 :

If $f (x)$ satisfies the relation $f (x) = e^{x} + \int_{0}^{1} (y + x e^{x}) f (y) d y,$ then $e + f (0)$ is equal to _______ . [2026]

Q 12 :

Let $I (x) = \int \frac{3 d x}{(4 x + 6) (\sqrt{4 x^{2} + 8 x + 3})}$ and $I (0) = \frac{\sqrt{3}}{4} + 20 .$ If $I (\frac{1}{2}) = \frac{a \sqrt{2}}{b} + c, where a, b, c \in ℕ, \gcd (a, b) = 1,$ then $a + b + c$ is equal to. [2026]

Q 13 :

If $\int ((\frac{1 - 5 \cos^{2} x}{\sin^{5} x \cos^{2} x})) d x = f (x) + C,$ where C is the constant of integration, then $f (\frac{π}{6}) - f (\frac{π}{4})$ is equal to [2026]

Q 14 :

$\int \frac{6 x^{10} + 4}{x^{3} \sqrt{x^{10} - 3 x^{4} - 1}} d x, x > 0 =$ _____________ (c is constant of integration)

Q 15 :

If $I = \int (x^{2} + 1) ({(x + 1) e^{x})}^{2} d x = A (f {(x))}^{2} + C,$ where C is constant of integration and $f (- 1) = \frac{2}{e},$ then $2 A + f (0)$ is

Q 16 :

$\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to:

Q 17 :

$\int \frac{d x}{\sin (x - a) \sin (x - b)}$ is equal to:

Q 18 :

$\int \frac{x^{3}}{x + 1} d x$ is equal to:

Q 19 :

$\int 7 e^{7 x + 5} d x =$

Q 20 :

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

Q 21 :

If $f' (x) = \frac{1}{x} + x$ and $f (1) = \frac{5}{2}$ , then $f (x) =$

Q 22 :

$\int 2^{x} 3^{x + 1} 4^{x + 2} d x =$

Q 23 :

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

Q 24 :

$\int x^{51} (\tan^{- 1} x + \cot^{- 1} x) d x =$

Q 25 :

The value of $\int \frac{1}{1 + \cos 8 x} d x$ is:

Q 26 :

By the inspection method, evaluate:

$\int (a u^{2} + b u + c) d u$

Q 27 :

Find the integral of $\sqrt{u} + \frac{1}{\sqrt{u}} .$

Q 28 :

$\int 4^{x} 3^{x} d x$ equals:

Q 29 :

Find the integral of $\sin u \sin 2 u \sin 3 u$ .

Q 30 :

$\int \tan^{- 1} \sqrt{x} d x$ is equal to:

Q 31 :

$\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$ is equal to:

Q 32 :

If $\int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = a {(1 + x^{2})}^{3 / 2} + b \sqrt{1 + x^{2}} + C,$ then:

Q 33 :

$\int \frac{1}{\log a} (a^{x} \cos a^{x}) d x =$

Q 34 :

Find the integral of $\frac{5 x^{4}}{\sqrt{x^{5} + 9}}$ :

Q 35 :

$\int \frac{\sec x d x}{\sqrt{\cos 2 x}} =$

Q 36 :

If $\int x^{6} \sin (5 x^{7}) d x = \frac{k}{5} \cos (5 x^{7}) + C, x \neq 0,$ then:

Q 37 :

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

Q 38 :

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

Q 39 :

$\int \frac{1}{1 + \cos^{2} x} d x =$

Q 40 :

Evaluate: $\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u$