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# 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)

July 23, 2017