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JEE Advanced 2008Paper 2

Exam: JEE Advanced
Year: 2008
Paper: Paper 2
Subject: Mathematics
Chapter: Indefinite Integration

JEE Advanced 2008 · Paper 2 · Mathematics · Indefinite Integration

Let I=∫ e^x e^ 4 x +e^ 2 x +1 d x, J=∫ e^ -x e^ -4 x +e^ -2 x +1 d...

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

Options

A $\frac{1}{2} \log \left|\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right|+C$
B $\frac{1}{2} \log \left|\frac{e^{2 x}+e^x+1}{e^{2 x}-e^x+1}\right|+C$
C $\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$
D $\frac{1}{2} \log \left|\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right|+C$

Correct answer

C. $\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$

Step-by-step solution

Since,

J=\int \frac{e^{3 x}}{1+e^{2 x}+e^{4 x}} d x

\begin{aligned} \thereforeJ-I & =\int \frac{\left(e^{3 x}-e^x\right)}{1+e^{2 x}+e^{4 x}}d x=\int \frac{\left(u^2-1\right)}{1+u^2+u^4}d u \\ & =\int \frac{\left(1-\frac{1}{u^2}\right)}{1+\frac{1}{u^2}+u^2}d u=\int \frac{\left(1-\frac{1}{u^2}\right)}{\left(u+\frac{1}{u}\right)^2-1}d u \\ & =\int \frac{d t}{t^2-1}\\ & =\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+C \\ & =\frac{1}{2} \log \left|\frac{u^2-u+1}{u^2+u+1}\right|+C=\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C \end{aligned}

$

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