<s>[INST] Evaluate the integral $I=\int_{0}^{\infty}\frac{\ln^3{x}}{(1+x^2)(1+x)^2}dx$ Find this integral
$$I=\int_{0}^{\infty}\dfrac{\ln^3{x}}{(1+x^2)(1+x)^2}dx$$
My try: let $x=\tan{t}$
then
$$I=\int_{0}^{\frac{\pi}{2}}\dfrac{\ln^3{\tan{t}}}{(1+\tan{t})^2}dt$$
I am unable to simplify after this. This problem is from QQ.
 [/INST] Filling in the steps of Mhenni Benghorbal's solution we have
The Mellin transform
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:160813a2}
\tag{2}
\end{equation}
where
\begin{equation}
f(x) = \frac{1}{(1+x^{2})(1+x)^{2}} = -\frac{1}{2}\frac{x}{x^{2}+1} + \frac{1}{2}\frac{1}{x+1} + \frac{1}{2}\frac{1}{(x+1)^{2}}
\label{eq:160813a3}
\tag{3}
\end{equation}
via partial fraction expansion.
Applying the Mellin transform, yields
\begin{align}
\mathcal{M}[f(x)](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(1+x^{2})(1+x)^{2}} \mathrm{d} x \\
& = -\frac{1}{2}\left[\frac{1}{2}\pi\sec\left(\frac{\pi}{2}s\right)\right] + \frac{1}{2}\pi\csc(\pi s) + \frac{1}{2} \mathrm{B}(s,2-s)
\label{eq:160813a4}
\tag{4}
\end{align}
Taking the 3rd derivative of equation \eqref{eq:160813a4} with respect to s and then taking $\lim s \to 1$ yields
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x & = -\frac{1}{2}\left( -\frac{7}{960} \pi^{4} \right) + \frac{1}{2} \left( -\frac{7}{60} \pi^{4} \right) + \frac{1}{2}0 \\
& = -\frac{7}{128} \pi^{4}
\label{eq:160813a5}
\tag{5}
\end{align}
Let us fill in the details. Handling the beta function first, we have
\begin{equation}
\mathrm{B}(s,2-s) = \frac{\Gamma(s)\Gamma(2-s)}{\Gamma(2)} = \Gamma(s)\Gamma(2-s)
\label{eq:160813a6}
\tag{6}
\end{equation}
To take derivatives, we note that
\begin{equation}
\frac{\mathrm d}{\mathrm d s} \Gamma(s) = \Gamma(s) \psi^{(0)}(s) \quad \mathrm{and} \quad \psi^{(n)}(s) = \frac{\mathrm{d}^{n}}{\mathrm{d} s^{n}} \psi^{(0)}(s)
\end{equation}
Where $\psi^{(n)}(s)$ is the polygamma function.
Taking the third derivative of equation \eqref{eq:160813a6} and letting $\lim s \to 1$ equals 0. Here we used
\begin{equation}
\psi^{(0)}(1) = -\gamma \quad \mathrm{and} \quad \psi^{(1)}(1) = \frac{\pi^{2}}{6}
\end{equation}
and fortunately $\Gamma^{(3)}(s) = -\Gamma^{(3)}(2-s)$ which leads to some cancellations.
Doing the same for the first two terms on the right hand side of equation \eqref{eq:160813a4}, we have to be careful. Each of them individually goes to $\infty$ as $\lim s \to 1$ but the $(s-1)^{-4}$ terms in the Laurent expansions about $s=1$ cancel. Here I used Wolfram Alpha. Doing it with the two terms combined yielded our final answer.
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