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

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
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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]

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(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]

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(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]

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(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]

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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

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

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

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

$We get, α = 2, β = 2, γ = 2, δ = 2$

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

Topic Question Set

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]

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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]

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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]

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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]

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Topic Question Set

Q 14 : If f(x)=∫1x1/4(1+x1/4)dx, f(0) = – 6, then f(1) is equal to : [2025] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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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]