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Home/JEE MAINS/mathematics/integrals
Curated PYQ Collection

Top 50 Most Repeated INTEGRALS PYQs | JEE MAINS

A curated collection of the most important questions from INTEGRALS, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

Concept Applied

Integrate term‑wise: \int 3x^2 dx = x^3, \int -4x dx = -2x^2, \int 1 dx = x. The antiderivative is x^3 - 2x^2 + x. Substituting the limits gives (8 - ...

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.$\frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$
B.$\frac{x}{2}e^{2x} + \frac{1}{4}e^{2x} + C$
C.$\frac{x}{2}e^{2x} - \frac{1}{2}e^{2x} + C$
D.$\frac{x}{2}e^{2x} + \frac{1}{2}e^{2x} + C$

Concept Applied

Choose $u = x$ (algebraic) and $dv = e^{2x}dx$ (exponential). Then $du = dx$ and $v = \frac{1}{2}e^{2x}$. Applying $\int u\,dv = uv - \int v\,du$ give...

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.$0$
B.$\pi$
C.$-\pi$
D.
...

Read Full Step-by-Step Solution →

Question #4

Practice Question

A.3\ln|x| - \ln|x+1| + C
B.3\ln|x| + \ln|x+1| + C
C.2\ln|x| + 3\ln|x+1| + C
D.\ln|x| + 2\ln|x+1| + C

Concept Applied

Decompose: (3x+2)/(x(x+1)) = A/x + B/(x+1). Solve: A=2, B=1 → integral = 2ln|x| + ln|x+1| + C....

Read Full Step-by-Step Solution →

Question #5

Practice Question

A.$\frac{\pi}{2}$
B.$\frac{\pi}{4}$
C.$\frac{\pi}{6}$
D.
$

Concept Applied

Using King property $f(x) \to f(\frac{\pi}{2}-x)$, sum of integrals is $\int_0^{\pi/2} 1 dx$....

Read Full Step-by-Step Solution →

Question #6

Practice Question

Concept Applied

After partial fraction decomposition, integral yields constant 3 on simplification....

Read Full Step-by-Step Solution →

Question #7

Practice Question

Concept Applied

Reduction formula: $ I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}I_{n-2} $. Coefficient of $ I_{n-2} $ is $ \frac{n-1}{n} $....

Read Full Step-by-Step Solution →

Question #8

Practice Question

A.$ -\frac{1}{2}\cos(2x) + C $
B.$ -2\cos(2x) + C $
C.$ \frac{1}{2}\cos(2x) + C $
D.$ 2\cos(2x) + C $

Concept Applied

Use substitution: Let $ u = 2x $, so $ du = 2dx $, hence $ dx = \frac{du}{2} $. Then $ \int \sin(2x)~dx = \frac{1}{2} \int \sin u~du = -\frac{1}{2}\co...

Read Full Step-by-Step Solution →

Question #9

Practice Question

Concept Applied

Odd function $\sin^3 x$ integrates to 0; $\cos x$ integrates to 2 over symmetric limits....

Read Full Step-by-Step Solution →

Question #10

Practice Question

Concept Applied

Let \(u = x^{2}+x+1\); then \(du = (2x+1)dx\). The integral becomes \(\int_{u=1}^{u=3} \sqrt{u}\,du = \frac{2}{3}u^{3/2}\big|_{1}^{3} = \frac{2}{3}(3\...

Read Full Step-by-Step Solution →

Question #11

Practice Question

Concept Applied

Antiderivative $=x^{3}+x^{2}$. Substituting limits: $(2^{3}+2^{2})-(0+0)=8+4=12$....

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.$\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + C$
B.$\tan^{-1}\left(\frac{x}{2}\right) + C$
C.$\frac{1}{2}\tan^{-1}(x) + C$
D.$\ln|x^2+4| + C$

Concept Applied

Standard integral: $\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\tan^{-1}(\frac{x}{a})+C$. Here $a=2$....

Read Full Step-by-Step Solution →

Question #13

Practice Question

Concept Applied

Antiderivative of 2x+5 is x²+5x. Evaluating from 1 to 3 gives (9+15)−(1+5)=24−6=18....

Read Full Step-by-Step Solution →

Question #14

Practice Question

Concept Applied

Integrate term‑wise: \int 3x^2 dx = x^3, \int -4x dx = -2x^2, \int 1 dx = x. The antiderivative is x^3 - 2x^2 + x. Substituting the limits gives (8 - ...

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.$\frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$
B.$\frac{x}{2}e^{2x} + \frac{1}{4}e^{2x} + C$
C.$\frac{x}{2}e^{2x} - \frac{1}{2}e^{2x} + C$
D.$\frac{x}{2}e^{2x} + \frac{1}{2}e^{2x} + C$

Concept Applied

Choose $u = x$ (algebraic) and $dv = e^{2x}dx$ (exponential). Then $du = dx$ and $v = \frac{1}{2}e^{2x}$. Applying $\int u\,dv = uv - \int v\,du$ give...

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.$0$
B.$\pi$
C.$-\pi$
D.
...

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.3\ln|x| - \ln|x+1| + C
B.3\ln|x| + \ln|x+1| + C
C.2\ln|x| + 3\ln|x+1| + C
D.\ln|x| + 2\ln|x+1| + C

Concept Applied

Decompose: (3x+2)/(x(x+1)) = A/x + B/(x+1). Solve: A=2, B=1 → integral = 2ln|x| + ln|x+1| + C....

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.$\frac{\pi}{2}$
B.$\frac{\pi}{4}$
C.$\frac{\pi}{6}$
D.
$

Concept Applied

Using King property $f(x) \to f(\frac{\pi}{2}-x)$, sum of integrals is $\int_0^{\pi/2} 1 dx$....

Read Full Step-by-Step Solution →

Question #19

Practice Question

Concept Applied

After partial fraction decomposition, integral yields constant 3 on simplification....

Read Full Step-by-Step Solution →

Question #20

Practice Question

Concept Applied

Reduction formula: $ I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}I_{n-2} $. Coefficient of $ I_{n-2} $ is $ \frac{n-1}{n} $....

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.$ -\frac{1}{2}\cos(2x) + C $
B.$ -2\cos(2x) + C $
C.$ \frac{1}{2}\cos(2x) + C $
D.$ 2\cos(2x) + C $

Concept Applied

Use substitution: Let $ u = 2x $, so $ du = 2dx $, hence $ dx = \frac{du}{2} $. Then $ \int \sin(2x)~dx = \frac{1}{2} \int \sin u~du = -\frac{1}{2}\co...

Read Full Step-by-Step Solution →

Question #22

Practice Question

Concept Applied

Odd function $\sin^3 x$ integrates to 0; $\cos x$ integrates to 2 over symmetric limits....

Read Full Step-by-Step Solution →

Question #23

Practice Question

Concept Applied

Let \(u = x^{2}+x+1\); then \(du = (2x+1)dx\). The integral becomes \(\int_{u=1}^{u=3} \sqrt{u}\,du = \frac{2}{3}u^{3/2}\big|_{1}^{3} = \frac{2}{3}(3\...

Read Full Step-by-Step Solution →

Question #24

Practice Question

Concept Applied

Antiderivative $=x^{3}+x^{2}$. Substituting limits: $(2^{3}+2^{2})-(0+0)=8+4=12$....

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.$\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + C$
B.$\tan^{-1}\left(\frac{x}{2}\right) + C$
C.$\frac{1}{2}\tan^{-1}(x) + C$
D.$\ln|x^2+4| + C$

Concept Applied

Standard integral: $\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\tan^{-1}(\frac{x}{a})+C$. Here $a=2$....

Read Full Step-by-Step Solution →

Question #26

Practice Question

Concept Applied

Antiderivative of 2x+5 is x²+5x. Evaluating from 1 to 3 gives (9+15)−(1+5)=24−6=18....

Read Full Step-by-Step Solution →

Question #27

Practice Question

Concept Applied

Integrate term‑wise: \int 3x^2 dx = x^3, \int -4x dx = -2x^2, \int 1 dx = x. The antiderivative is x^3 - 2x^2 + x. Substituting the limits gives (8 - ...

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.$\frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$
B.$\frac{x}{2}e^{2x} + \frac{1}{4}e^{2x} + C$
C.$\frac{x}{2}e^{2x} - \frac{1}{2}e^{2x} + C$
D.$\frac{x}{2}e^{2x} + \frac{1}{2}e^{2x} + C$

Concept Applied

Choose $u = x$ (algebraic) and $dv = e^{2x}dx$ (exponential). Then $du = dx$ and $v = \frac{1}{2}e^{2x}$. Applying $\int u\,dv = uv - \int v\,du$ give...

Read Full Step-by-Step Solution →

Question #29

Practice Question

A.$0$
B.$\pi$
C.$-\pi$
D.
...

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.3\ln|x| - \ln|x+1| + C
B.3\ln|x| + \ln|x+1| + C
C.2\ln|x| + 3\ln|x+1| + C
D.\ln|x| + 2\ln|x+1| + C

Concept Applied

Decompose: (3x+2)/(x(x+1)) = A/x + B/(x+1). Solve: A=2, B=1 → integral = 2ln|x| + ln|x+1| + C....

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.$\frac{\pi}{2}$
B.$\frac{\pi}{4}$
C.$\frac{\pi}{6}$
D.
$

Concept Applied

Using King property $f(x) \to f(\frac{\pi}{2}-x)$, sum of integrals is $\int_0^{\pi/2} 1 dx$....

Read Full Step-by-Step Solution →

Question #32

Practice Question

Concept Applied

After partial fraction decomposition, integral yields constant 3 on simplification....

Read Full Step-by-Step Solution →

Question #33

Practice Question

Concept Applied

Reduction formula: $ I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}I_{n-2} $. Coefficient of $ I_{n-2} $ is $ \frac{n-1}{n} $....

Read Full Step-by-Step Solution →

Question #34

Practice Question

A.$ -\frac{1}{2}\cos(2x) + C $
B.$ -2\cos(2x) + C $
C.$ \frac{1}{2}\cos(2x) + C $
D.$ 2\cos(2x) + C $

Concept Applied

Use substitution: Let $ u = 2x $, so $ du = 2dx $, hence $ dx = \frac{du}{2} $. Then $ \int \sin(2x)~dx = \frac{1}{2} \int \sin u~du = -\frac{1}{2}\co...

Read Full Step-by-Step Solution →

Question #35

Practice Question

Concept Applied

Odd function $\sin^3 x$ integrates to 0; $\cos x$ integrates to 2 over symmetric limits....

Read Full Step-by-Step Solution →

Question #36

Practice Question

Concept Applied

Let \(u = x^{2}+x+1\); then \(du = (2x+1)dx\). The integral becomes \(\int_{u=1}^{u=3} \sqrt{u}\,du = \frac{2}{3}u^{3/2}\big|_{1}^{3} = \frac{2}{3}(3\...

Read Full Step-by-Step Solution →

Question #37

Practice Question

Concept Applied

Antiderivative $=x^{3}+x^{2}$. Substituting limits: $(2^{3}+2^{2})-(0+0)=8+4=12$....

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.$\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + C$
B.$\tan^{-1}\left(\frac{x}{2}\right) + C$
C.$\frac{1}{2}\tan^{-1}(x) + C$
D.$\ln|x^2+4| + C$

Concept Applied

Standard integral: $\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\tan^{-1}(\frac{x}{a})+C$. Here $a=2$....

Read Full Step-by-Step Solution →

Question #39

Practice Question

Concept Applied

Antiderivative of 2x+5 is x²+5x. Evaluating from 1 to 3 gives (9+15)−(1+5)=24−6=18....

Read Full Step-by-Step Solution →

Question #40

Practice Question

Concept Applied

Integrate term‑wise: \int 3x^2 dx = x^3, \int -4x dx = -2x^2, \int 1 dx = x. The antiderivative is x^3 - 2x^2 + x. Substituting the limits gives (8 - ...

Read Full Step-by-Step Solution →

Question #41

Practice Question

A.$\frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$
B.$\frac{x}{2}e^{2x} + \frac{1}{4}e^{2x} + C$
C.$\frac{x}{2}e^{2x} - \frac{1}{2}e^{2x} + C$
D.$\frac{x}{2}e^{2x} + \frac{1}{2}e^{2x} + C$

Concept Applied

Choose $u = x$ (algebraic) and $dv = e^{2x}dx$ (exponential). Then $du = dx$ and $v = \frac{1}{2}e^{2x}$. Applying $\int u\,dv = uv - \int v\,du$ give...

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.$0$
B.$\pi$
C.$-\pi$
D.
...

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.3\ln|x| - \ln|x+1| + C
B.3\ln|x| + \ln|x+1| + C
C.2\ln|x| + 3\ln|x+1| + C
D.\ln|x| + 2\ln|x+1| + C

Concept Applied

Decompose: (3x+2)/(x(x+1)) = A/x + B/(x+1). Solve: A=2, B=1 → integral = 2ln|x| + ln|x+1| + C....

Read Full Step-by-Step Solution →

Question #44

Practice Question

A.$\frac{\pi}{2}$
B.$\frac{\pi}{4}$
C.$\frac{\pi}{6}$
D.
$

Concept Applied

Using King property $f(x) \to f(\frac{\pi}{2}-x)$, sum of integrals is $\int_0^{\pi/2} 1 dx$....

Read Full Step-by-Step Solution →

Question #45

Practice Question

Concept Applied

After partial fraction decomposition, integral yields constant 3 on simplification....

Read Full Step-by-Step Solution →

Question #46

Practice Question

Concept Applied

Reduction formula: $ I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}I_{n-2} $. Coefficient of $ I_{n-2} $ is $ \frac{n-1}{n} $....

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.$ -\frac{1}{2}\cos(2x) + C $
B.$ -2\cos(2x) + C $
C.$ \frac{1}{2}\cos(2x) + C $
D.$ 2\cos(2x) + C $

Concept Applied

Use substitution: Let $ u = 2x $, so $ du = 2dx $, hence $ dx = \frac{du}{2} $. Then $ \int \sin(2x)~dx = \frac{1}{2} \int \sin u~du = -\frac{1}{2}\co...

Read Full Step-by-Step Solution →

Question #48

Practice Question

Concept Applied

Odd function $\sin^3 x$ integrates to 0; $\cos x$ integrates to 2 over symmetric limits....

Read Full Step-by-Step Solution →

Question #49

Practice Question

Concept Applied

Let \(u = x^{2}+x+1\); then \(du = (2x+1)dx\). The integral becomes \(\int_{u=1}^{u=3} \sqrt{u}\,du = \frac{2}{3}u^{3/2}\big|_{1}^{3} = \frac{2}{3}(3\...

Read Full Step-by-Step Solution →

Question #50

Practice Question

Concept Applied

Antiderivative $=x^{3}+x^{2}$. Substituting limits: $(2^{3}+2^{2})-(0+0)=8+4=12$....

Read Full Step-by-Step Solution →
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