TITLE: Prove that $\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\ln{3}$
QUESTION [5 upvotes]: We wish to prove that 

$$I=\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\ln{3}\tag1$$

$$1-e^{3x}=(1-e^x)(1+e^x+e^{2x})\tag2$$
$$1+e^{3x}=(1+e^x)(1-e^x+e^{2x})\tag3$$
Sub $(2)$ and $(3)$ into $(1)\rightarrow (4)$
$$I=\int_{0}^{\infty}\left(2\cdot{1\over 1+e^x+e^{2x}}+{1\over 1-e^x+e^{2x}}\right)dx\tag5$$
Any hint, please, I am unable to continue.
$$I=\int_{0}^{1}\left({2u\over u^2+u+1}+{u\over u^2-u+1}\right)du=I_1+I_2\tag6$$
Respectively.
Apply formula (17) to $(6)$
Hence
$$I_1=\ln{3}-{2\sqrt3\over 3}\tan^{-1}\left({3\over \sqrt3}\right)$$
$$I_2={2\over \sqrt3}\tan^{-1}\left({\sqrt3 \over 3}\right)$$
Hence $I=\ln{3}$

REPLY [3 votes]: $$I=\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\int_{0}^{\infty}\frac{1-e^x}{1-e^{3x}}dx+2\int_{0}^{\infty}\frac{1-e^{4x}}{1-e^{6x}}dx$$
$$I=\int_{0}^{\infty}\frac{1}{e^x(1+e^x+e^{2x})}e^xdx+\int_{0}^{\infty}\frac{1+e^{2x}}{e^{2x}(1+e^{2x}+e^{4x})}2e^{2x}dx$$
Set $u=e^x$ in the first integral and $u=e^{2x}$ in the second one, so we have
$$I=\int_{1}^{\infty}\frac{u+2}{u(1+u+u^2)}du=\int_{1}^{\infty}\left(\frac{2}{u}-\frac{2u+1}{1+u+u^2}\right)du=\ln\left(\frac{u^2}{1+u+u^2}\right)\Big{|}_{1}^{\infty}=\color{red}{\ln 3}$$