Let $$I = \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx,\,\,J = \int {{{{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}}dx.} } $$ Then for an arbitrary constant $$C$$, the value of $$J -I$$ equals :

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

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$ is equal to

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

The integral $\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x$ equals (for some arbitrary constant $$K$$)

$-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$

JEE Advanced · Quantitative Aptitude

JEE Advanced Integration By Substitution Questions with Answers and Solutions

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JEE Advanced Integration By Substitution practice

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Let

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

for

n \geq 2

and

g(x)=\underbrace{(f o f o \ldots . o f)}_{f \text { occurs } n \text { times }}(x)

. Then

\int x^{n-2} g(x) d x

equals :

Choose one answer to reveal the solution.

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JEE Advanced Integration By Substitution practice

What visitors can use on this page

JEE Advanced Integration By Substitution practice FAQ

Prepare together, not alone

JEE Advanced Integration By Substitution practice

What visitors can use on this page

JEE Advanced Integration By Substitution practice FAQ