Complex analysis contour integral I am working on the integral $\displaystyle\int_0^{\infty}\frac{\log(x)}{x^2-1}$. I see it done here $\int_0^\infty\frac{\log x dx}{x^2-1}$ with a hint. but I am wondering if it is possible to integrate on a different contour, the keyhole. I know that we could set the contour integral equal to the residue at $z = i$ which is $\dfrac{\pi}{2}$ since it is a simple pole. I also know that the integral around the large circle will be zero.

I'm not sure the semicircle is exactly what you want.  The standard keyhole contour works quite well here, as I will show below.
Note that the singularity at $x=1$ is removable in this integral.  We evaluate this integral by  appealing to the residue theorem; consider 
$$\oint_{C} dz \frac{\log^2{z}}{z^2-1}$$
where $C$ is a keyhole contour with respect to the positive real axis.  By integrating around this contour and noting that the integrand vanishes sufficiently fast as the radius of the circular section of $C$ increases without bound, we get
$$\oint_{C} dz \frac{\log^2{z}}{z^2-1} = -i 4 \pi  \int_0^{\infty} dx \frac{\log{x}}{x^2-1} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2-1}$$
This is equal to, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand of the complex integral within $C$.  As the only pole is at $z=-1$, we see that
$$\begin{align}\oint_{C} dz \frac{\log^2{z}}{z^2-1} &= i 2 \pi \frac{\log^2{(-1)}}{2 (-1)} \\ &= i 2 \pi  \frac{\pi^2}{2}\end{align}$$ 
Now, the real part of the integral above is split into a Cauchy principal value and a piece indented about the singularity at $x=1$.  The Cauchy principal value is zero:
$$\begin{align}PV \int_0^{\infty} dx \frac{1}{x^2-1} &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_{1+\epsilon}^{\infty} dx \frac{1}{x^2-1}\right]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_0^{1/(1+\epsilon)} \left (-\frac{dx}{x^2} \right ) \frac{1}{(1/x^2)-1} \right ]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} - \int_0^{1-\epsilon}  \frac{dx}{x^2-1} \right ] \\ &= 0\end{align}$$
The indent in the contour, however, produces a contribution; let $x=1+\epsilon e^{i \phi}$ and $\phi \in [\pi,0]$:
$$4 \pi^2 i \epsilon \int_{-\pi}^0 d\phi \frac{e^{i \phi}}{2 \epsilon e^{i \phi}} = i \frac{\pi}{2} 4 \pi^2$$
so that
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = i 2 \pi \frac{\pi^2}{2} - i \frac{\pi}{2} 4 \pi^2 = -i 2 \pi \frac{\pi^2}{2}$$
Therefore
$$\int_0^{\infty} dx \frac{\log{x}}{x^2-1} = \frac{\pi^2}{4}$$
