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: 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:
$J=4 \times \dfrac{\pi^2}{12}+4 \times \dfrac{\pi^2}{6}=\pi^2$
$I=8G+\pi^2$
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: This problem is awful (at least to me !). I was wondering if a CAS could do it; it did and I am sure that you will enjoy the result $$\frac{G_{3,3}^{3,2}\left(1\left|
\begin{array}{c}
 0,\frac{3}{4},1 \\
 0,0,0
\end{array}
\right.\right)}{\Gamma \left(\frac{1}{4}\right)}$$ in which appears the Meijer G function (I am sure you are happy to know that !). The numerical value of this nice expression is $17.19732915$.  
If I may ask, where did you get this monster from ?
Added to my answer
Since I produced the above monster, I let it !   
If you make a change of variable such that $x+1=e^{4y}$, you have something which is more pleasant. The result of the integral is $8 C+\pi ^2$, $C$ being the Catalan number.
A: With $y^4=x+1$, Maple gets the antiderivative
$$
V(y)=2\,\ln  \left( y+i \right) \pi +2\,\ln  \left( y-i \right) \pi -4\,
\ln  \left( 1+y \right) \ln  \left( y \right) -4\,i{\rm Li}_2
 \left( 1-iy \right) +4\,i{\rm Li}_2 \left( 1+iy \right) -4\,{
\rm Li}_2 \left( -y \right) -4\,{\rm Li}_2 \left( 1-y
 \right)
$$
Now $V(0)=-8 \;\mathrm{Catalan} + \pi^2/3$ and $\lim_{y\to\infty} V(y) = 4\pi^2/3$, so subtract to get our answer $8 \;\mathrm{Catalan} + \pi^2$.
