Using the Residue theorem to evaluate $\int_{0}^{\infty}\frac{\ln(x)}{(x+1)^{3}}dx$. I need help to evaluate the integral 
\begin{align*}
\int_{0}^{\infty}\frac{\ln(x)}{(x+1)^{3}}dx
\end{align*}
using the Residue theorem. 
I think that I could consider the contour integral
\begin{align*}
\oint_{C}\frac{\ln(z)}{(z+1)^{3}}dz,
\end{align*}
where $C$ is the quarter circle in the upper-right quadrant. But I'm not sure how to continue. 
Could someone help me?
 A: Take the function $$f\left(z\right)=\frac{\log^{2}\left(z\right)}{\left(1-z\right)^{3}}$$ and the branch of the logarithm corresponding to $-\pi<\arg\left(z\right)\leq\pi$. Take the keyhole contour and define $\Gamma$ and $\gamma$ respectively the large circumference of radius $R$ and the small circumference of radius $\rho$ (see here).
It is not difficult to prove that the integrals over the circumferences vanish so $$2\pi i\underset{z=1}{\textrm{Res}}\left(f\left(z\right)\right)=\int_{0}^{\infty}\frac{\log^{2}\left(-x+i\epsilon\right)}{\left(x-i\epsilon+1\right)^{3}}dx-\int_{0}^{\infty}\frac{\log^{2}\left(-x-i\epsilon\right)}{\left(x+i\epsilon+1\right)^{3}}dx$$ $$\stackrel{\epsilon\rightarrow0}{\rightarrow}\int_{0}^{\infty}\frac{\left(\log\left(x\right)+i\pi\right)^{2}}{\left(x+1\right)^{3}}dx-\int_{0}^{\infty}\frac{\left(\log\left(x\right)-i\pi\right)^{2}}{\left(x+1\right)^{3}}dx$$ $$=4\pi i\int_{0}^{\infty}\frac{\log\left(x\right)}{\left(x+1\right)^{3}}dx$$ and since $$\underset{z=1}{\textrm{Res}}\left(f\left(z\right)\right)=-1$$ we get $$\int_{0}^{\infty}\frac{\log\left(x\right)}{\left(x+1\right)^{3}}dx=\color{red}{-\frac{1}{2}}.$$
We can also use real methods. Note that $$I\left(a\right)=\int_{0}^{\infty}\frac{x^{a}}{\left(x+1\right)^{3}}dx=B\left(a+1,-a+2\right),\,-1<a<2$$ where $B(x,y)$ is the beta function so $$I'\left(a\right)=\int_{0}^{\infty}\frac{\log\left(x\right)x^{a}}{\left(x+1\right)^{3}}dx=\left(\psi\left(a+1\right)-\psi\left(2-a\right)\right)B\left(a+1,-a+2\right)$$ where $\psi\left(x\right)$ is the digamma function hence $$I=\lim_{a\rightarrow0}\left[\left(\psi\left(a+1\right)-\psi\left(2-a\right)\right)B\left(a+1,-a+2\right)\right]=\color{blue}{-\frac{1}{2}}.$$
