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
Answer & explanation
Correct answer: option 3
We have,
$J-I=\int \frac{e^{3 x}}{e^{4 x}+e^{2 x}+1}-\frac{e^x}{e^{4 x}+e^{2 x}+1} d x$
$\Rightarrow J-I =\int \frac{e^{2 x}-1}{e^{4 x}+e^{2 x}+1} d\left(e^x\right)$
$\Rightarrow J-I =\int \frac{t^2-1}{t^4+t^2+1} d t$, where $t=e^x$
$\Rightarrow J-I=\int \frac{1}{\left(t+\frac{1}{t}\right)^2-1} d\left(t-\frac{1}{t}\right)$
$\Rightarrow J-I=\frac{1}{2} \log \left(\frac{t+\frac{1}{t}-1}{t+\frac{1}{t}+1}\right)+C$
$\Rightarrow J-I=\frac{1}{2} \log \left(\frac{t^2-t+1}{t^2+t+1}\right)+C=\frac{1}{2} \log \left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)+C$