Finding $\int_{0}^{\infty} \frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)} dx$

Question

I want to find the following integral: $$\int_{0}^{\infty} \frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)} dx$$ I have been trying to apply simple substitutions like $u=e^x$, to simplify the integral without any luck. I simplified it to

$$\int_{0}^{\infty} \frac{1}{x(e^x+1)}-\frac{1}{x(e^{3x}+1)}dx$$ which seems to have increased my problems.

Wolfram Alpha gives the answer to be 0.5493... and the graph looks like a Gaussian distribution.

Somehow, I feel that this integral is very non-trivial due to the 'troublesome' $x$ in the denominator.

Answer 1

By the Frullani's theorem we have $$\int_{0}^{+\infty}\frac{1}{x}\left(\frac{1}{e^{x}+1}-\frac{1}{e^{3x}+1}\right)dx=\color{red}{\frac{\log\left(3\right)}{2}}$$ taking $f(x):=\frac{1}{e^{x}+1},\,a=1,\,b=3$.

Answer 2

According to your work, we have that $$\begin{align}\int_{0}^{\infty} \frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)} dx&= \lim_{t\to 0}\left(\int_{t}^{\infty} \frac{dx}{x(e^x+1)}-\int_{t}^{\infty}\frac{dx}{x(e^{3x}+1)}\right)\\ &= \lim_{t\to 0}\left(\int_{t}^{\infty} \frac{dx}{x(e^x+1)}-\int_{3t}^{\infty}\frac{dy}{y(e^{y}+1)}\right)\\ &= \lim_{t\to 0}\left(\int_{t}^{3t} \frac{dx}{x(e^x+1)}\right)\\ &=\lim_{t\to 0}\left(\int_{t}^{3t} \frac{dx}{x(2+O(x))}\right) =\frac{\log\left(3\right)}{2}. \end{align}$$