Definite integral by Malmsten's formula I was able to evaluate the definite integral
$$I=\int_0^{\infty}\frac{e^x-1}{x(e^{2x}+1)}dx=\log\Gamma\left(\frac{1}{4}\right)-\log\Gamma\left(\frac{3}{4}\right)-\log\Gamma\left(\frac{1}{2}\right)$$
by using the Malmsten's formula
$$\log(\Gamma(z))=\int_0^{\infty}\left(\frac{e^{-zt}-e^{-t}}{1-e^{-t}}-(z-1)e^{-t}\right)\frac{dt}{t}$$
going backwards from the answer:
$$\log(\Gamma(\frac{1}{4}))=\int_0^{\infty}\left(\frac{e^{-\frac{1}{4}t}-e^{-t}}{1-e^{-t}}-\frac{3}{4}e^{-t}\right)\frac{dt}{t}$$
$$-\log(\Gamma(\frac{3}{4}))=\int_0^{\infty}\left(\frac{-e^{-\frac{3}{4}t}+e^{-t}}{1-e^{-t}}+\frac{1}{4}e^{-t}\right)\frac{dt}{t}$$
$$-\log(\Gamma(\frac{1}{2}))=\int_0^{\infty}\left(\frac{-e^{-\frac{1}{2}t}+e^{-t}}{1-e^{-t}}+\frac{1}{2}e^{-t}\right)\frac{dt}{t}$$
By adding,
$$I=\int_0^{\infty}\frac{e^{-\frac{1}{4}t}-e^{-\frac{3}{4}t}-e^{-\frac{1}{2}t}+e^{-t}}{1-e^{-t}}\frac{dt}{t}$$
and with $t=4x$,
$$I=\int_0^{\infty}\frac{e^{3x}-e^{x}-e^{2x}+1}{e^{4x}-1}\frac{dx}{x}=\int_0^{\infty}\frac{(e^{2x}-1)(e^{x}-1)}{(e^{2x}-1)(e^{2x}+1)}\frac{dx}{x}=\int_0^{\infty}\frac{e^x-1}{x(e^{2x}+1)}dx.$$
I just learned this formula today. I wonder if it's possible to evaluate this integral without using this formula. Also, if I didn't know the answer I couldn't evaluate this integral!
Any idea or methods are wellcome. Thanks in advance.
 A: This is not a full solution, but it is the best I got.
Let $I$ be the original integral. Then
$$
\eqalign{
I &= \int_{0}^{\infty}\frac{e^{x}-1}{x\left(e^{2x}+1\right)}dx \cr
&= \int_{0}^{\infty}\frac{\left(e^{x}-1\right)e^{x}}{x\left(e^{2x}+1\right)e^{x}}dx \cr
&= \int_{1}^{\infty}\left(\frac{x-1}{x\left(1+x^{2}\right)\ln\left(x\right)}\right)dx \cr
&= \int_{1}^{\infty}\int_{0}^{1}\frac{x^{t}}{x\left(1+x^{2}\right)}dtdx \cr
&= \int_{1}^{\infty}\int_{0}^{1}\frac{x^{t-1}}{1+x^{2}}dtdx \cr
&= \int_{1}^{\infty}\int_{0}^{1}\sum_{n=0}^{\infty}\frac{t^{n}\ln\left(x\right)^{n}}{xn!+x^{3}n!}dtdx \cr
&= \sum_{n=0}^{\infty}\frac{1}{n!\left(n+1\right)}\int_{1}^{\infty}\frac{\ln\left(x\right)^{n}}{x+x^{3}}dx \cr
&= \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}-1}{4^{n}\left(n+1\right)}\sum_{k=1}^{\infty}\frac{1}{k^{n+1}}\right) \cr
&= \frac{1}{2}\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{2^{n}-1}{4^{n}k^{n+1}\left(n+1\right)} \cr
&= \frac{1}{2}\ln{(2\pi)} - 2\ln{\Gamma(3/4)}.
}
$$
According to WolframAlpha, my steps are correct. However, I am clueless about how to prove
$$\int_{1}^{\infty}\frac{\ln^{n}\left(x\right)}{x+x^{3}}dx\ =\ \frac{\left(2^{n}-1\right)n!}{2^{\left(2n+1\right)}}\zeta{(n+1)},$$
and the last equality, too. If someone could fill in the blanks, I would appreciate it.
A: I think this is a more natural solution:
$\begin{align}
\int_0^{\infty}\frac{e^x-1}{x(e^{2x}+1)}dx &=\int_0^{\infty}\frac{e^x-1}{x}\frac{e^{-2x}}{1+e^{-2x}}dx \\
&=\int_0^{\infty}\sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}\sum_{k=1}^{\infty}(-1)^{k+1}e^{-2kx}dx\\
&=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(n+1)!}\int_0^{\infty}x^ne^{-2kx}dx \\
&=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(n+1)!}\frac{n!}{2^{n+1}k^{n+1}}\\
&=\sum_{n=0}^{\infty}\frac{1}{2^{n+1}(n+1)}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n+1}}\\
&=\frac{\ln2}{2}+\sum_{n=1}^{\infty}\frac{1}{2^{n+1}(n+1)}\eta(n+1)\\
&=\frac{\ln2}{2}+\sum_{n=1}^{\infty}\frac{1}{2^{n+1}(n+1)}(1-2^{-n)}\zeta(n+1)\\
&=\frac{\ln2}{2}+\sum_{n=1}^{\infty}\frac{2^n-1}{2^{2n+1}(n+1)}\sum_{k=1}^{\infty}\frac{1}{k^{n+1}}\\
&=\frac{\ln2}{2}+\frac{1}{2}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{2^n-1}{4^{n}k^{n+1}(n+1)}\\
&=\frac{\ln2}{2}+\ln\Gamma(\frac{1}{2})-2\ln\Gamma(\frac{3}{4})\\
&=\frac{\ln(2\pi)}{2}-2\ln\Gamma(\frac{3}{4})\\
\end{align}$
