Compute $\int_0^\infty \frac{\ln x}{(1+x)^3}\,\mathrm{d}x$ 
Compute $$\int_0^\infty \frac{\ln x}{(1+x)^3}\,\mathrm{d}x$$

Well by comparison test the integral is convergent. I tried to use residue theorem, with the positive real axis being the branch cutting line, but it did not work. Any hint will be appreciated.
 A: The answer is $-1/2$.
Indeed, let the considered integral be denoted by $I$. Clearly we have
$$\eqalignno{
I&=\int_0^1\frac{\ln x}{(1+x)^3}dx+\int_1^\infty\frac{\ln x}{(1+x)^3}dx\cr
&=\int_0^1\frac{\ln x}{(1+x)^3}dx+\int_0^1\frac{\ln (1/x)}{(1+1/x)^3}\frac{1}{x^2}dx\cr
&=\int_0^1\frac{(1-x)\ln x}{(1+x)^3}dx\cr
&=\left[\frac{x}{(1+x)^2}\ln x\right]_0^1-\int_0^1\frac{dx}{(1+x)^2}\cr
&=\left[\frac{1}{1+x}\right]_0^1=-\frac{1}{2}.&\square
}$$
A: I imagine you're having an issue with cancellation.
Let $ \displaystyle f(z) = \frac{\log^{{\color{red}{2}}}z}{(1+z)^{3}}$ and integrate around a keyhole contour with the branch cut for the logarithm placed along the positive real axis.
Then 
$$ \begin{align} \int_{0}^{\infty} \frac{\log^{2} x}{(1+x)^{3}} \ dx + \int_{\infty}^{0} \frac{(\log x + 2 \pi i)^{2}}{(1+x)^{3}} \ dx &= 2 \pi i \ \text{Res} [f(z),e^{\pi i}] \\&=2 \pi i \lim_{z \to e^{\pi i}} \frac{1}{2!} \frac{d^{2}}{dz^{2}} \log^{2}(z) \\&=  2\pi i \lim_{z \to e^{\pi i}} \frac{1 - \log z}{z^{2}} \\&= 2 \pi i \ \frac{1-\pi i}{1} \\&=2 \pi ^{2} + 2 \pi i. \end{align}$$
So we have 
$$ - 4 \pi i \int_{0}^{\infty} \frac{\log x}{(1+x)^{3}} + 4 \pi^{2} \int_{0}^{\infty} \frac{1}{(1+x)^{3}} \ dx   = 2 \pi ^{2} + 2 \pi i .$$
Equating the imaginary parts on both sides of the equation,
$$ - 4 \pi i \int_{0}^{\infty} \frac{\log x}{(1+x)^{3}} \ dx = 2 \pi i$$
which implies 
$$ \int_{0}^{\infty} \frac{\log x}{(1+x)^{3}} \ dx = - \frac{1}{2}.$$
