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…

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
  1. $\frac{1}{2} \log \left|\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right|+C$
  2. $\frac{1}{2} \log \left|\frac{e^{2 x}+e^x+1}{e^{2 x}-e^x+1}\right|+C$
  3. $\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$
  4. $\frac{1}{2} \log \left|\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right|+C$

Solution

Since, $J=\int \frac{e^{3 x}}{1+e^{2 x}+e^{4 x}} d x$ $ \begin{aligned} \therefore J-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} $

Asked in: JEE Advanced 2008 (Paper 2)

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