How to integrate $\int_1^{\infty}\frac{\sqrt{x-1}}{(1+x)^2}dx$ using contour integration? How to integrate  using contour integration?
$$\int_1^{\infty}\frac{\sqrt{x-1}}{(1+x)^2}dx$$
I was putting $y = x-1$ then $\frac{dy}{dx}$= $1-0$, ${dy}={dx} $   then i get $$\int_1^{\infty}\frac{\sqrt{y}}{(2+y)^2}dy$$
I don't know how to take it from here. I would appreciate if someone could help me and give me some hints or the solution. Thanks.

I won't prefer a contour here, making the substitution $y=x-1$ we have,
$$ \displaystyle \int_1^\infty \dfrac{\sqrt{x-1}}{(1+x)^2}\; dx = \int_0^\infty \dfrac{\sqrt{y}}{(2+y)^2}\; dy$$
Next substitute $y=u^2$ , 
$$\displaystyle I = 2\int_0^\infty \dfrac{u^2}{(u^2+2)^2}\; du $$
Using partial fractions this will be equal to,
$$ \displaystyle \begin{align} I &= 2\int_0^\infty \dfrac{du}{u^2+2}-\int_0^\infty \dfrac{4\; du}{(u^2+2)^2} \\ &= \dfrac{\pi}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\int_0^{\pi/2}  (1+\cos 2\theta)\; d\theta \\ &=\dfrac{\pi}{\sqrt{2}}-\dfrac{\pi}{2\sqrt{2}} \\ &= \dfrac{\pi}{2\sqrt{2}} \end{align}$$
Footnotes :
1) $\displaystyle \int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a} \tan^{-1}\dfrac{x}{a}$
If you integrate it then,
$\displaystyle \int_0^{+\infty} \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\times \dfrac{\pi}{2}$
2) $\displaystyle \int_0^{+\infty} \dfrac{4\; du}{(u^2+2)^2}$
We make the substitution $u=\sqrt{2}\tan \theta$ , this means $du=\sqrt{2}\sec^2 \theta d\theta$ .The new limits for $\theta$ would be according to 
$ \displaystyle \theta = \tan^{-1}\dfrac{u}{\sqrt{2}}$ .The new limits are thus from $0\to\dfrac{\pi}{2}$ since $\tan^{-1}\dfrac{+\infty}{\sqrt{2}}=\dfrac{\pi}{2}$ & $\tan^{-1}\dfrac{0}{\sqrt{2}}=0$ . The integral now reads :
$$\displaystyle \begin{align}\int_0^{+\infty} \dfrac{4\; du}{(u^2+2)^2} &= 4\int_0^{\pi/2} \dfrac{\sqrt{2}\sec^2 \theta}{(2(\tan^2 \theta+1))^2} \; d\theta \\ &= 4\sqrt{2}\int_0^{\pi/2} \dfrac{\sec^2 \theta}{4(\sec^2 \theta)^2}\; d\theta \\ &= \sqrt{2} \int_0^{\pi/2} \dfrac{\sec^2 \theta}{\sec^4 \theta} \; d\theta \\ &= \sqrt{2} \int_0^{\pi/2} \cos^2 \theta \; d\theta \\ &= \dfrac{\sqrt{2}}{2}\int_0^{\pi/2} 2\cos^2 \theta \; d\theta \\ &= \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} (1+\cos 2\theta)\; d\theta \\ &= \dfrac{1}{\sqrt{2}} \int_0^{\pi/2} d\theta + \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} \cos 2\theta \; d\theta \\ &= \dfrac{\pi}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}\left[\sin 2\theta\right]_0^{\pi/2} \\ &= \dfrac{\pi}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}(\sin \pi - \sin 0) \\ &= \dfrac{\pi}{2\sqrt{2}} \end{align} $$
