I know how to solve this integral without contour integration. The answer to the integral is
$$\int^{\infty }_{0}\frac{\ln( z)}{( z+1)^{\alpha}} \,dx=\frac{H_{\alpha-2}}{1-\alpha} ,\; \alpha>1,$$ where $H_k$ is the Harmonic number function, and $\alpha$ is not limited to the integers.
To solve it without contour integration I used Feynman's technique. But I know that there are branch points at $z=0, -1$ in the integrand. So it looks like I should be able to solve it with contour integration. But I couldn't. I tried using this contour
but was unable to get the wanted integral when integrating around the contour. It makes me think that I would need to insert another natural log into the integrand to add another branch point (so that with the argument limiting, I would get my integral back), But I already have two branch points so I'm not sure if that would work. Thanks in advance for the help!
Let $n\in \mathbb{N}$, $n\ge 2$ and let $f(z)=\frac{\log^2(z)}{(z+1)^n}$. Note that $f$ has a branch point at $z=0$ and a simple pole at $z=-1$. Cut the plane from $z=0$ to $z=\infty$, along the positive real axis.
The residue theorem guarantees that
$$\begin{align} \int_0^\infty \frac{\log^2(x)-(\log(x)+i2\pi)^2}{(x+1)^n}\,dx&=2\pi i \text{Res}\left(\frac{\log^2(z)}{(z+1)^n}, z=-1\right)\\\\ &=\frac{2\pi i}{(n-1)!}\lim_{z\to -1}\frac{d^{n-1}\log^2(z)}{dz^{n-1}}\\\\ &=\frac{4\pi i}{(n-1)!}\lim_{z\to -1}\frac{d^{n-2}\left(z^{-1}\log(z)\right)}{dz^{n-2}}\\\\ &=\frac{4\pi i}{(n-1)!}\lim_{z\to -1}\left(\frac{(-1)^n(n-2)!\log(z)}{z^{n-1}}\\+\sum_{k=1}^{n-2}\binom{n-2}{k}\frac{d^k\log(z)}{dx^k}\frac{d^{n-2-k}x^{-1}}{dz^{n-2-k}}\right)\\\\ &=\frac{4\pi i}{n-1}\left(-i\pi+\sum_{k=1}^{n-2}\frac1k\right)\tag1 \end{align}$$
Simplifying $(1)$ recovers the expected result for $n\in \mathbb{N}$, $n\ge2$
$$\int_0^\infty\frac{\log(x)}{(x+1)^n}\,dx=\frac{H_{n-2}}{1-n}$$