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.
 A: Here is a closed form

$$I=\int_{0}^{\infty}\dfrac{\ln^3{x}}{(1+x^2)(1+x)^2}dx =  -\frac{7}{128} \pi^4 \sim -5.327059668.$$

You can use this technique.
A: What I propose only aims at simplifying the evaluation. Consider the following contour integral:
$$\oint_\gamma \frac{\log^4(z)}{(1+z^2)(1\color{red}{-}z)^2}dz=2\pi i \sum \text{Res}f(z)$$
Notice the minus in $\color{red}{\text{red}}$ in the denominator instead of the plus in the original function; you will soon find out why I did this. Taking $\gamma$ to be the keyhole contour  about the negative axis, and setting $z=-x-i\epsilon$ for the lower line and $z=-x+i\epsilon$ for the upper, you get:
$$-\int_\infty^0 \frac{\log^4(-x+i\epsilon)}{(1+(-x+i\epsilon)^2)(1\color{red}{-}(-x+i\epsilon))^2}dx-\int_0^\infty \frac{\log^4(-x-i\epsilon)}{(1+(-x-i\epsilon)^2)(1\color{red}{-}(-x-i\epsilon))^2}dx$$
$$\int_0^\infty \frac{(\log(x)+i\pi)^4}{(1+x^2)(1\color{blue}{+}x)^2}dx-\int_0^\infty \frac{(\log(x)-i\pi)^4}{(1+x^2)(1\color{blue}{+}x)^2}dx$$
$$\int_0^\infty\frac{8\pi i(\log^3(z)-\pi^2\log(z))}{(1+x^2)(1+x)^2}dx=8\pi i\left(I-\pi^2J\right)$$
$I$ is the integral that you're looking for, while $J$ is the integral containing $\log(z)$ which can be evaluated using the same trick (keyhole contour). It turns out to be $$J=-\frac{\pi^2}{16}$$
Furthermore, the residues in the first equation above can be found in a straightforward way, the sum turns out to be:
$$\sum\text{Res}f(z)=\frac{\pi^4}{32}$$ so that: $$8\pi i\left(I+\frac{\pi^4}{16}\right)=2\pi i \frac{\pi^4}{32}$$
Solving for $I$ we get:
$$\boxed{\color{blue}{I=-\frac{7}{128}\pi^4}}$$
A: Here is an elementary way. First note that
$$\int_1^{\infty} \dfrac{\ln^3(x) dx}{(1+x^2)(1+x)^2} = \int_1^0 \dfrac{-\ln^3(x)}{(1+1/x^2)(1+1/x)^2} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{-x^2 \ln^3(x)}{(1+x^2)(1+x)^2}dx$$
Hence, your integral is
$$I = \int_0^1 \dfrac{(1-x^2) \ln^3(x)}{(1+x^2)(1+x)^2}dx = \underbrace{\int_0^1 \dfrac{\ln^3(x)}{1+x}dx}_J - \overbrace{\int_0^1 \dfrac{x\ln^3(x)}{1+x^2}dx}^K$$
We have
$$J = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^k \ln^3(x)dx = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac6{(k+1)^4} = -\dfrac7{120} \pi^4 \tag{$\star$}$$
We have
$$K = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^{2k+1} \ln^3(x)dx = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac3{8(k+1)^4} = -\dfrac7{1920} \pi^4 \tag{$\dagger$}$$
Hence,
$$\boxed{\color{red}{I = J-K = -\dfrac7{128}\pi^4}}$$
Where we used the following facts
$$\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(k+1)^4} = -\dfrac78 \zeta(4)$$
$$\zeta(4) = \dfrac{\pi^4}{90}$$
to simplify $(\star)$ and $(\dagger)$.
A: 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.
A: I think what Mhenni Benghorbal suggested is this:
\begin{align*}
I\left(a\right): & =\int_{0}^{\infty}\frac{x^{a}}{\left(1+x^{2}\right)\left(1+x\right)^{2}}dx=\frac{\pi}{\sin\pi a}\left(\frac{1}{2}\cos\frac{\pi a}{2}+\frac{a-1}{2}\right).
\end{align*}
Then 
\begin{align*}
\left(\frac{d}{da}\right)^{m}I\left(0\right) & =\int_{0}^{\infty}\frac{\left(\ln x\right)^{m}}{\left(1+x^{2}\right)\left(1+x\right)^{2}}dx\\
 & =\left(\frac{d}{da}\right)^{m}\Big|_{a=0}\,\,\frac{\pi}{\sin\pi a}\left(\frac{1}{2}\cos\frac{\pi a}{2}+\frac{a-1}{2}\right).
\end{align*}
