Evaluate $\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$ using Gamma and Beta functions I need to evaluate the following integral, using Gamma and Beta functions:
$$\int_0^\infty\frac{\sqrt[5]{x}}{(x^2+1)(x+1)^2}dx$$
I tried to use partial fractions, but I am not sure if that is correct, since doing that I get that the integral is divergent.
Since
$\frac{1}{\left(x^2+1\right)\left(x+1\right)^2}=\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{\left(x+1\right)^2}\right)-\frac{1}{2}\left(\frac{x}{x^2+1}\right)$
Then,
$\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{\left(x^2+1\right)\left(x+1\right)^2}dx=\frac{1}{2}\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{x+1}dx}+\frac{1}{2}\int_{0}^{\infty}{\frac{\sqrt[5]{x}}{\left(x+1\right)^2}dx}-\frac{1}{2}\int_{0}^{\infty}{\frac{x\sqrt[5]{x}}{x^2+1}dx}}$
But you can see that the first of this new integrals is divergent, and we only need one integral to be divergent to say that the whole integral diverges, but it is not true since the original integral is convergent, so there must be something wrong with this procedure.
 A: We can make your approach work by modifying it slightly. Consider the integral
$$I(s)=\int_0^\infty\frac{x^{s-1}}{(1+x^2)(1+x)^2}\,\mathrm dx$$
instead. Applying your partial fraction expansion, an integral representation of the beta function $\text{B}(x,\,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\,\mathrm dt$, and the reflection formula for the gamma function $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$ tells us that
\begin{align*}
I(s) &= \frac{1}{2}\int_0^\infty\frac{x^{s-1}}{1+x}\,\mathrm dx+\frac{1}{2}\int_0^\infty\frac{x^{s-1}}{(1+x)^2}\,\mathrm dx-\frac{1}{2}\int_0^\infty\frac{x^s}{1+x^2}\,\mathrm dx \\
&= \frac{1}{2}\text{B}(s,\,1-s)+\frac{1}{2}\text{B}(s,\,2-s)-\frac{1}{4}\text{B}\left(\frac{s+1}{2},\,\frac{1-s}{2}\right) \\
&= \frac{\pi}{2}(2-s)\csc(\pi s)-\frac{\pi}{4}\sec\left(\frac{\pi s}{2}\right).
\end{align*}
This last expression agrees with I(s) for $0<\Re(s)<4$. Setting $s=6/5$ yields
$$I\left(\frac{6}{5}\right)=\frac{\pi\phi}{2}-\frac{4\pi}{5}\sqrt{\frac{2}{5}+\frac{\phi}{5}}$$
where $\phi$ is the golden ratio.
A: Substitute  $x=\tan u$
$$\begin{align}
  & =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\tan u}}{{{(\tan u+1)}^{2}}}du} \\ 
 & =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\frac{\sin u}{\cos u}}}{{{(\frac{\sin u}{\cos u}+1)}^{2}}}du} \\ 
 & =\int_{0}^{\pi /2}{\frac{\sqrt[5]{\sin u}}{\sqrt[5]{\cos u}}\frac{{{\cos }^{2}}u}{{{(\sin u+\cos u)}^{2}}}du} \\ 
 & =\int_{0}^{\pi /2}{{{\sin }^{1/5}}u{{\cos }^{9/5}}u\frac{1}{{{(1+2\sin u\cos u)}^{2}}}du} \\ 
\end{align}$$
Using  $$\frac{1}{{{(1+y)}^{2}}}=\sum\nolimits_{n=0}^{\infty }{{{(-1)}^{n}}(1+n){{y}^{n}}}$$
$$\begin{align}
  & =\int_{0}^{\pi /2}{\left( \sum\nolimits_{n=0}^{\infty }{{{(-2)}^{n}}(1+n){{\sin }^{1/5+n}}u{{\cos }^{9/5+n}}u} \right)du} \\ 
 & =\sum\nolimits_{n=0}^{\infty }{{{(-2)}^{n}}(1+n)\left( \int_{0}^{\pi /2}{{{\sin }^{1/5+n}}u{{\cos }^{9/5+n}}udu} \right)} \\ 
 & =\sum\nolimits_{n=0}^{\infty }{{{(-2)}^{n}}(1+n)\left( \frac{1}{2}Beta \left( n/2+3/5,n/2+7/5 \right) \right)} \\ 
\end{align}$$
Is it possible to continue ??????
