Q:

Integration do not know how to proceed: $\int_0^{\infty}\frac{\ln(x+1)}{x\cdot(x+1)^\frac14} dx$

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

A:

$$I(n)=\int_0^\infty(x+1)^n\frac{dx}x\iff\int_0^{\infty}\frac{\ln(x+1)}{x\cdot(x+1)^\frac14} dx=I'\bigg(-\frac14\bigg)$$
Let $t=\dfrac1{x+1}$ , and recognize the expression of the beta function in the new integral, then use the reflection formula, but do not compute its value, since it will be divergent ! Rather, derive it directly with regard to n. The expression of the derivative will converge. But it will require knowledge of the polygamma function, as well as of generalized harmonic numbers. In the end, the entire expression will eventually simplify to the beautiful $\pi^2+8G$, where G is Catalan's constant. Hope this helps ! I was also thinking of an alternative, by expanding $\ln(1+x)$ into its well-known Taylor series, but I just don't see a way around the $\sqrt[4]{1+x}$ in the denominator. Maybe someone else can chime in and shed some more light into this situation, or maybe even develop a better or more elegant approach.

A:

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:
$J=4 \times \dfrac{\pi^2}{12}+4 \times \dfrac{\pi^2}{6}=\pi^2$
$I=8G+\pi^2$

