TITLE: Integration do not know how to proceed: $\int_0^{\infty}\frac{\ln(x+1)}{x\cdot(x+1)^\frac14} dx$
QUESTION [5 upvotes]: Integration 0 to infinity 
$$\int_0^\infty\frac{\ln(x+1)}{x\cdot(x+1)^\frac14}dx$$
How to proceed ?
Answer : $8C+\pi^2$ 
Procedure wanted

REPLY [4 votes]: First, assume $\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2}=\dfrac{\pi^2}{6}$
Then:  
$\displaystyle \dfrac{\pi^2}{6}=\sum_{n=1}^{+\infty} \dfrac{1}{n^2}=\sum_{n=1}^{+\infty} \dfrac{1}{(2n)^2}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=\dfrac{\pi^2}{24}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}$
Therefore:
 $\displaystyle \sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=\dfrac{\pi^2}{8}$ (1)
Let $\displaystyle I=\int_0^{+\infty}\dfrac{\ln(x+1)}{x (x+1)^\frac14}dx$
Perform change of variable $u=(x+1)^\frac14$:
$\displaystyle I=\int_1^{+\infty} \dfrac{16x^2\log(x)}{x^4-1}dx$
Therefore:
$\displaystyle I=\int_1^{+\infty} \dfrac{8\log(x)}{x^2-1}dx+\int_1^{+\infty} \dfrac{8\log(x)}{x^2+1}dx$
The last integral equals $8G$ where $G$ is the Catalan's constant.
Let $\displaystyle J=\int_1^{+\infty} \dfrac{8\log(x)}{x^2-1}dx$
Perform change of variable $u=\dfrac{x-1}{x+1}$:
$\displaystyle J=\int_0^1 4\log\Big(\dfrac{1+x}{1-x}\Big)\dfrac{dx}{x}=4\Big(\int_0^1 \dfrac{\log(1+x)dx}{x}-\int_0^1 \dfrac{\log(1-x)dx}{x}\Big)$
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=\int_0^1  \Big(\sum_{n=0}^{+\infty} \dfrac{(-1)^n x^n}{n+1}\Big) dx=\sum_{n=0}^{+\infty} \Big( \int_0^1\dfrac{(-1)^nx^n}{n+1}dx  \Big)$
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{(n+1)^2}$
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=\sum_{n=1}^{+\infty} \dfrac{(-1)^{2n+1}}{(2n)^2}+\sum_{n=0}^{+\infty} \dfrac{(-1)^{2n+2}}{(2n+1)^2}=-\dfrac{\pi^2}{24}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}$
Apply (1) :
$\displaystyle \int_0^1 \dfrac{\log(1+x)dx}{x}=-\dfrac{\pi^2}{24}+\dfrac{\pi^2}{8}=\dfrac{\pi^2}{12}$
$\displaystyle -\int_0^1 \dfrac{\log(1-x)dx}{x}=\int_0^1 \Big(\sum_{n=0}^{+\infty} \dfrac{x^n}{n+1}\Big) dx=\sum_{n=0}^{+\infty}  \Big(\int_0^1\dfrac{x^n}{n+1}dx  \Big)=\sum_{n=0}^{+\infty} \dfrac{1}{(n+1)^2}=\sum_{n=1}^{+\infty} \dfrac{1}{n^2}= \dfrac{\pi}{6}$
Therefore:
$J=4 \times \dfrac{\pi^2}{12}+4 \times \dfrac{\pi^2}{6}=\pi^2$
$I=8G+\pi^2$