Compute $\int_{0}^{\infty} \frac{\sqrt{x}}{(1+x)^2} dx$

Question

I couldn’t find the right transformation for this integral. Any help would be appreciated.

$$\int_{0}^{\infty} \frac{\sqrt{x}}{(1+x)^2} dx$$

Answer 1

Set $y=x+1$, so you get $$\int_1^\infty\frac{(y-1)^{1/2}}{y^2}\,dy.$$ Now set $t=1/y$ to give $$\int_0^1(1/t-1)^{1/2}\,dt=\int_0^1t^{-1/2}(1-t)^{1/2}=B(1/2,3/2).$$

Answer 2

$y=\frac x {1+x}$ transforms the integral to $\int_0^{1} y^{1/2}(1-y)^{-1/2} \, dy$.

Answer 3

Alternative approach. Note that by integration by parts we can easily evaluate the integral directly, $$\int_{0}^{\infty} \frac{\sqrt{x}}{(1+x)^2} dx= -\left[\frac{\sqrt{x}}{(1+x)}\right]_{0}^{\infty} +\frac{1}{2}\int_{0}^{\infty} \frac{1}{\sqrt{x}(1+x)} dx=0+ \left[\arctan(\sqrt{x})\right]_{0}^{\infty}=\frac{\pi}{2}.$$

Answer 4

The first thing that may come to mind is to try the substitution $u = \sqrt{x}$. Then $x = u^2$ and $dx = 2u\ du$. The integral becomes

$$ \int_0^\infty \frac{2u^2}{(1+u^2)^2}\ du $$

Then let $u = \tan t$ to get a trig integral

$$ \int_{0}^{\pi/2} \frac{2\tan^2 t}{\sec^4 t}\sec^2 t\ dt = 2\int_0^{\pi/2} \sin^2 t\ dt = \frac{\pi}{2} $$