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?

Let $u=\sqrt{x-1}$, $du=\frac{1}{2\sqrt{x-1}}$, then
$$I=2\int_0^\infty\frac{u^2}{(u^2+2)^2}du$$
We can apply partial fractions here. 
$$=2\int_0^\infty\left(\frac{1}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$
$$=2\int_0^\infty\frac{1}{u^2+2}du-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
The first integrand is almost $\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\frac{u}{\sqrt{2}}$.
$$I=\sqrt 2\int_0^\infty\frac{1}{s^2+1}ds-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
$$=\frac{\pi}{\sqrt 2}-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$
Now to tackle the second integral we can use a trig sub. 
Let $u=\sqrt{2}\tan (t)$, $du=\sqrt 2 \sec^2(t)dt$.
$$I=\frac{\pi}{\sqrt 2}-4\sqrt 2 \int_0^{\pi/2}\frac{\sec ^2 t}{4\sec^4t}dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2\int _0^{\pi/2}\cos^2(t) dt$$
$$=\frac{\pi}{\sqrt 2}-\sqrt 2 \int_0^{\pi/2}\left(\frac{1}{2}\cos(2t)+\frac{1}{2} \right)dt$$ 
If we substitute $v=2t$ and split the integral up we get
$$I=\frac{\pi}{\sqrt 2}-\frac{1}{2\sqrt 2} \int_0^\pi \cos(v)dv-\frac{1}{\sqrt 2}\int_0^{\pi/2}1dt$$
The first integral clearly goes to 0 and the second integral becomes $\frac{\pi}{2\sqrt 2}$.
Therefore 
$$I=\frac{\pi}{2\sqrt 2}$$
