A Tricky Integral

Let $$ I=\int e^x/(e^{4x}+e^{2x}+1) dx$$ $$ J= \int e^{-x}/(e^{-4x}+e^{-2x}+1)dx$$. Find the value of \(J-I\).

Solution:

$$ I=\int e^x/(e^{4x}+e^{2x}+1) dx$$

$$J= \int e^{-x}/(e^{-4x}+e^{-2x}+1)dx$$

Let \(e^x\)=\(z\)

$$ J-I=\int\frac{e^x(e^{2x-1})}{e^{4x}+e^{2x}+1}dx=\int\frac{z^2-1}{z^4+z^2+1}dz
$$

$$ =\frac{1}{2}ln\frac{(e^x+e^-x-1)}{(e^x+e^-x+1)}+c$$ ( where c is a constant of integration)

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