The value of $\mathop \smallint \nolimits_1^2 \,\,\frac{{dx}}{{x\left( {1 + {x^4}} \right)}}$ is

$\frac{1}{4}\,\,\log \,\,\frac{{17}}{{32}}$

$\frac{1}{4}\,\,\log \,\,\frac{{17}}{2}$

$\smallint \,\,{\left( {\frac{{x + 2}}{{x + 4}}} \right)^2}{e^x}\,\;dx$ is equal to

${e^x}\left( {\frac{{x + 2}}{{x + 4}}} \right) + c$

${e^x}\left( {\frac{{x - 2}}{{x + 4}}} \right) + c$

$\left( {\frac{{2x{e^x}}}{{x + 4}}} \right) + c$

$\smallint \frac{{\sin \,\,2x}}{{1\,\, + \,\,{{\cos }^2}\,\,x}}\,\,dx$ is equal to

$ - \frac{1}{2}\,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}x} \right) + c$

$2\,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}x} \right) + c$

$\frac{1}{2}\,\,\log \,\,\left( {1\,\, + \,\,\cos \,\,\,2x} \right) + c$

$\smallint \frac{{x + 2}}{{\left( {{x^2} + 3x + 3} \right)\,\,\sqrt {x + 1} }}dx$ is equal to

$\frac{1}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {3\left( {x + 1} \right)} }}} \right)$

$\frac{2}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {x + 1} }}} \right)$

If $\int \frac{1}{1-\sin ^4 x} d x=\frac{1}{2} \tan x+A \tan ^{-1}\{f(x)\}+C$, then

$A=\sqrt{2}$ and $f(x)=\sqrt{2} \tan x$

$A=-\sqrt{2}$ and $f(x)=\sqrt{2} \tan x$

If $f(x)=\int \frac{\tan ^3 x}{2+\tan ^2 x} d x$, then

$f(x)=\log \left|\frac{2-\cos ^2 x}{\sin x}\right|+C$

$f(x)=\log \left|\frac{2+\cos ^2 x}{\sin x}\right|+C$

The integral $\int {{{\sec }^{2/3}}x\,{\rm{cose}}{{\rm{c}}^{4/3}}x\,dx} $ is equal to (here $C$ is a constant of integration)

$ - \frac{3}{4}{\tan ^{ - 4/3}}x + C$

$\smallint \,\,\frac{{{x^4} - \,\,1}}{{{x^2}\,\sqrt {{x^4} + \,\,{x^2} + 1} }}\,\,dx$ is equal to

$\frac{x}{{\sqrt {{x^4} + {x^2} + 1} }} + C$

$\frac{{2x}}{{\sqrt {{x^4} + {x^2} + 1} }} + C$

$\frac{{\sqrt {{x^4} + {x^2} + 1} }}{{2x}} + C$

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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

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481 Easy

481 of 995

Application +4 | -1

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

View answer and explanation

Correct option: C

\frac{2}{5} \tan ^5 \sqrt{x}+c

Option C is correct because the integral can be solved using substitution.

482 Medium

482 of 995

Analyzing +4 | -1

Let

f\left( x \right) = \frac{x}{{{{\left( {1 + {x^n}} \right)}^{1/n}}}}

for

n\,\, \ge 2

and

\mathrm{g}\,\,\left( x \right) = \left( {fofo\,\, \ldots of} \right)\,\,\left( x \right)

g(x) = \underbrace {(fofo....of)}_{n\,\,times}\,\,(x)

Then,

\int \,\,{x^{n - 2}}\,\,\mathrm{g}\left( x \right)\,\,dx

equals

View answer and explanation

Correct option: A

\frac{1}{{n\left( {n - 1} \right)}}\,\,{\left( {1 + n{x^n}} \right)^{1 - \frac{1}{n}}}\,\, + \,\,c

Option A is correct because by finding the composite function g(x) and then integrating x^(n-2)*g(x), we arrive at the given expression.

483 Easy

483 of 995

Application +4 | -1

The value of the integral

\int \limits_2^4 \,\,\frac{{\sqrt {{x^2} - \,\,4} }}{{{x^4}}}\,\,dx

, is

View answer and explanation

Correct option: B

\frac{{\sqrt 3 }}{{32}}

Option B (

\frac{{\sqrt 3 }}{{32}}

) is correct because it is the result of evaluating the definite integral using trigonometric substitution.

484 Hard

484 of 995

Analyzing +4 | -1

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

equals

View answer and explanation

Correct option: B

-2 \tan^{-1} \left(\frac{\sqrt{1+x-x^2}+1}{x}+1\right)+c

Option B (

-2 \tan^{-1} (\frac{{\sqrt{{1+x-x^2}}+1}}{{x}}+1)+c

) is correct because after applying Euler's substitution and simplifying, the integral evaluates to this form.

485 Easy

485 of 995

Application +4 | -1

\int e^x\left(1+\tan x+\tan ^2 x\right) d x=

View answer and explanation

Correct option: C

e^x \tan x+c

Option C is correct because the integral simplifies to

e^x \tan x + c

by recognizing the form

\int e^x [f(x) + f'(x)] dx = e^x f(x) + c

486 Medium

486 of 995

Analyzing +4 | -1

\int \frac{\sin ^4 x}{\cos x} d x=\frac{1}{2} \log _e\left(\frac{1+\sin x}{1-\sin x}\right)-g(x)+c

where

g(x)

equals

View answer and explanation

Correct option: A

\frac{1}{3} \sin ^3 x+\sin x

Option A (

\left[-1,-\frac{3}{4}\right]

) is correct because the range of the expression

\frac{1}{4}(\sin 2 \theta)^2-1

falls within this interval.

487 Medium

487 of 995

Application +4 | -1

The value of

\int \nolimits_1^2 \,\,\frac{{dx}}{{x\left( {1 + {x^4}} \right)}}

View answer and explanation

Correct option: B

\frac{1}{4}\,\,\log \,\,\frac{{32}}{{17}}

Option B (

\frac{1}{4}\,\log \,\frac{{32}}{{17}}

) is correct because after substituting

x^4 = t

and using partial fractions, we can evaluate the definite integral.

488 Medium

488 of 995

Application +4 | -1

\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+C

(where

C

is a constant of integration), then the ordered pair

(A, B)

is equal to

View answer and explanation

Correct option: A

(-2,-1)

Option A ((-2,-1)) is correct because it is the result of integrating the given function and comparing the coefficients with the given form.

489 Easy

489 of 995

Analyzing +4 | -1

\int \,\,{\left( {\frac{{x + 2}}{{x + 4}}} \right)^2}{e^x}\,\;dx

is equal to

View answer and explanation

Correct option: A

{e^x}\left( {\frac{x}{{x + 4}}} \right) + c

Option A (

{e^x}\left( {\frac{x}{{x + 4}}} \right) + c

) is correct because the integral can be solved by manipulating the integrand and using integration by parts.

490 Easy

490 of 995

Application +4 | -1

\int \frac{{\sin \,\,2x}}{{1\,\, + \,\,{{\cos }^2}\,\,x}}\,\,dx

is equal to

View answer and explanation

Correct option: D

c\,\, - \,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}\,\,x} \right)

Option D,

c\,\, - \,\,\log \,\,\left( {1\,\, + \,\,{{\cos }^2}\,\,x} \right)

, is correct because it is the result of integrating the given trigonometric function using a simple substitution.

491 Medium

491 of 995

Application +4 | -1

\int \frac{{x + 2}}{{\left( {{x^2} + 3x + 3} \right)\,\,\sqrt {x + 1} }}dx

is equal to

View answer and explanation

Correct option: B

\frac{2}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {3\left( {x + 1} \right)} }}} \right)

Option B,

\frac{2}{{\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{{\sqrt {3\,\,\left( {x + 1} \right)} }}} \right)

, is correct because it is the result of the given integral after applying a suitable substitution and trigonometric simplification.

492 Easy

492 of 995

Application +4 | -1

\int f(x) d x=\phi(x)

i.e.

\phi

is a function such that

\phi^{\prime}(x)

=f(x)

, then

\int x^9 f\left(x^5\right) d x

is equal to

View answer and explanation

Correct option: C

\frac{1}{5} x^5 \phi\left(x^5\right)-\int x^4 \phi\left(x^5\right) d x+C

Option C (

\frac{1}{5} x^5 \phi(x^5)-\int x^4 \phi(x^5) d x+C

) is correct because it is derived by using substitution and integration by parts.

493 Medium

493 of 995

Application +4 | -1

\int \log 2 x d x

View answer and explanation

Correct option: D

x \log 2 x-x+c

Option D (

x \log 2 x-x+c

) is correct because it is the result of integrating

\log 2x

using integration by parts.

494 Medium

494 of 995

Analyzing +4 | -1

For

0 < x < 1

, let

f(x)=\lim \limits_{{n \rightarrow \infty}}(1+x)\left(1+x^2\right)\left(1+x^4\right) \cdots\left(1+x^{2^n}\right)

then

\int \frac{f(x)}{1-x} \log _e x d x

equals

View answer and explanation

Correct option: B

-\log _e\left(\frac{x}{1-x}\right)+\frac{\log _e x}{1-x}+c

Option B is correct because the integral can be solved by simplifying the limit, then using integration by parts.

495 Easy

495 of 995

Application +4 | -1

If the integral

\int \frac{5 \tan x}{\tan x-2} d x=x+a \log

|\sin x-2 \cos x|+k

, then

a

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because it represents the value of 'a' obtained by comparing the given integral with the provided expression after solving the integral.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

496 Easy

496 of 995

Application +4 | -1

\int \frac{\left(x^2-1\right) \mathrm{d} x}{\left(x^4+3 x^2+1\right) \tan ^{-1} \frac{x^2+1}{x}}=

\log \left|\tan ^{-1} f(x)\right|+C

then

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because after dividing the numerator and denominator by

x^2

and using appropriate substitutions, the integral simplifies to

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

497 Easy

497 of 995

Analyzing +4 | -1

\int \frac{1}{1-\sin ^4 x} d x=\frac{1}{2} \tan x+A \tan ^{-1}\{f(x)\}+C

, then

View answer and explanation

Correct option: A

A=\frac{1}{2 \sqrt{2}}

and

f(x)=\sqrt{2} \tan x

Option A is correct because after simplifying the integral using trigonometric identities and partial fractions, the given values for A and f(x) are obtained.

498 Medium

498 of 995

Analyzing +4 | -1

f(x)=\int \frac{\tan ^3 x}{2+\tan ^2 x} d x

, then

View answer and explanation

Correct option: B

f(x)=\log \left|\frac{2-\sin ^2 x}{\cos x}\right|+C

The correct answer is

f(x)=\log \left|\frac{{2-\sin ^2 x}}{{\cos x}}\right|+C

(Option B) because the integral simplifies to this expression after trigonometric substitution and simplification.

499 Easy

499 of 995

Application +4 | -1

\int\limits_{{-1}}^1 x^2 \sin 2 x d x=

View answer and explanation

Correct option: B

-\frac{1}{2} x^2 \cos 2 x+\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+c

Option B,

-\frac{1}{2} x^2 \cos 2 x+\frac{1}{2} x \sin 2 x+\frac{1}{4} \cos 2 x+c

, is the correct antiderivative, but the question asks for the definite integral from -1 to 1 of an odd function, which is 0.

500 Medium

500 of 995

Application +4 | -1

\int \frac{{dx}}{{\sin \,\,x\,\,\cos \,\,x}}

is equal to

View answer and explanation

Correct option: B

\log \,\,|\tan \,\,x| + \,c

Option B is correct because the integral of

\frac{{dx}}{{\sin \,x\,\cos \,x}}

simplifies to

\log \,|\tan \,x| + \,c

using trigonometric identities and integration rules.

501 Easy

501 of 995

Understanding +4 | -1

\int \,\,\frac{{f'\,\,\left( x \right)}}{{\sqrt {f\,\,\left( x \right)} }}\,\,dx = ...\,\, + \,\,c\,;\,\,f\left( x \right) \ne 0

View answer and explanation

Correct option: B

2\,\,\sqrt {f\,\,\left( x \right)}

Option B (

2\,\,\sqrt {f\,\,\left( x \right)}

) is correct because the integral of

\frac{{f'\,\,\left( x \right)}}{{\sqrt {f\,\,\left( x \right)} }}

2\sqrt{f(x)} + c

502 Easy

502 of 995

Understanding +4 | -1

\int \frac{{f'\left( x \right)}}{{f\left( x \right)\,\,\log \,\,\left[ {f\left( x \right)} \right]}}\,\,dx

is equal to

View answer and explanation

Correct option: C

\log \,\,\left[ {\log \,\,f\left( x \right)} \right]\,\, + \,\,c

Option C (log[log f(x)] + c) is correct because it is the result of applying u-substitution twice, first with u = log(f(x)) and then with v = log(u).

503 Easy

503 of 995

Application +4 | -1

\int \frac{\sin ^4 x}{\cos ^8 x} d x=a \tan ^7 x+b \tan ^5 x+C

, then

View answer and explanation

Correct option: A

7 a=5 b

Option B is correct because after simplifying the integral using trigonometric identities and integration, we find 5a = 7b.

504 Easy

504 of 995

Application +4 | -1

The integral

\int {{{\sec }^{2/3}}x\,\mathrm{cose}{\mathrm{c}^{4/3}}x\,dx}

is equal to (here

C

is a constant of integration)

View answer and explanation

Correct option: B

- 3{\tan ^{ - 1/3}}x + C

Option B (

- 3{\tan ^{ - 1/3}}x + C

) is correct because the integral evaluates to this expression after substitution and simplification.

505 Easy

505 of 995

Analyzing +4 | -1

Let

I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x,

Then, for an arbitrary constant

C

, the value of

J-I

equals

View answer and explanation

Correct option: C

\frac{1}{2} \log \left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)+C

Option C (

\frac{1}{2} \log \left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)+C

) is correct because after simplifying the integrals and using substitution, the result matches this option.

506 Easy

506 of 995

Analyzing +4 | -1

The value of

I=\int \frac{d x}{\left(a+b x^2\right) \sqrt{b-a x^2}}

, is

View answer and explanation

Correct option: A

\frac{1}{\sqrt{a\left(a^2+b^2\right)}} \tan ^{-1}\left(\frac{x \sqrt{a^2+b^2}}{a \sqrt{b-a x^2}}\right)+C

Option A (

\frac{1}{\sqrt{a(a^2+b^2)}} \tan ^{-1}\left(\frac{x \sqrt{a^2+b^2}}{a \sqrt{b-a x^2}}\right)+C

) is correct because the integral simplifies to this inverse tangent form after trigonometric substitution.

507 Easy

507 of 995

Application +4 | -1

\int \tan ^3 2 x \sec 2 x d x=

View answer and explanation

Correct option: A

\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c

Option A (

\frac{{1}}{{6}} \sec ^3 2 x-\frac{{1}}{{2}} \sec 2 x+c

) is correct because the integral can be solved by substitution and using trigonometric identities.

508 Easy

508 of 995

Application +4 | -1

\int \frac{e^x d x}{\sqrt{1-e^{2 x}}}=

View answer and explanation

Correct option: B

-\cos ^{-1}\left(e^x\right)+c

Option B (

-\cos ^{-1}\left(e^x\right)+c

) is correct because it is the result of a u-substitution and applying the standard integral of

\frac{1}{\sqrt{1-t^2}}

509 Easy

509 of 995

Analyzing +4 | -1

\int \,\,\frac{{{x^4} - \,\,1}}{{{x^2}\,\sqrt {{x^4} + \,\,{x^2} + 1} }}\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{{\sqrt {{x^4} + {x^2} + 1} }}{x} + C

Option B (

\frac{{\sqrt {{x^4} + {x^2} + 1} }}{x} + C

) is correct because after simplifying and integrating the given expression, the result matches this form.

510 Medium

510 of 995

Application +4 | -1

Let

f(x)=x+\sin x

. Suppose

g

denotes the inverse function of

f

. The value of

g^{\prime}\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)

has the value equal to

View answer and explanation

Correct option: C

2-\sqrt{2}

Option C (

2-\sqrt{2}

) is correct because it is the value of the derivative of the inverse function

g

evaluated at the given point, calculated using the derivative of the original function

f

Revision notes and key takeaways

Review INDEFINITE INTEGRATION in your textbook or class notes, then use this set to check your understanding.

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