Evaluate $ \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x} $. I need to solve the following integral:
$$
I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.
$$
Wolfram Alpha gives the answer as $ \dfrac{\pi}{2 \sqrt{2}} $.
I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?

I think I have a different answer. 
Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then
$$\begin{align}
\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx 
&= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\
&= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du
\end{align}$$
Now let $u = \frac{1}{t}$, so that $du = \frac{-1}{t^2} dt$. Continuing.
$$\begin{align}
\phantom{\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx} 
&= \frac{1}{\sqrt{2}} \int_0^1 t^2 (1-t)^\frac{1}{2} \frac{1}{\sqrt{t}} \frac{1}{t^2} dt \\
&= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{-1}{2}} (1-t)^\frac{1}{2} dt \\
&= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{2} - 1} dt \\
&= \frac{1}{\sqrt{2}} B ( \frac{1}{2} . \frac{3}{2} ) \\
&= \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{3}{2})} {\sqrt{2}\Gamma(2)} \\
&= \frac{\pi}{2\sqrt{2}}.
\end{align}$$
Original image of work by hand.
