A problem asked by Jon Helge Angeltveit

Evaluate $\int \frac{\sqrt{x}}{4 -x^2} \; dx $

Answer:

Substitute $u =\sqrt{x}$, and we have $du = \frac{1}{2\sqrt{x}} \; dx $ or, $dx = 2u \; du$ and

$\int \frac{\sqrt{x}}{4 -x^2} \; dx = \int \frac{2u^2}{4-u^4} \; du$

$= \int \frac{1}{2 - u^2} \; du - \int \frac{1}{2 + u^2} \; du $

$= \frac{1}{2\sqrt{2}} \int \frac{1}{\sqrt{2} - u} \; du + \frac{1}{2\sqrt{2}} \int \frac{1}{\sqrt{2} + u } \; du - \int \frac{1}{2 + u^2} \; du $

$= \frac{-1}{2\sqrt{2}} ln|u - \sqrt{2}| + \frac{1}{2\sqrt{2}} ln|u + \sqrt{2}| - \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{u}{\sqrt{2}} \right) + C$

$= \frac{1}{2\sqrt{2}} ln\biggr|\frac{u + \sqrt{2}}{u - \sqrt{2}}\biggr| - \frac{1}{\sqrt{2}} \tan^{-1} \left( \sqrt{\frac{x}{2}} \right) + C$

$= \frac{1}{2\sqrt{2}} ln\biggr|\frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} - \sqrt{2}}\biggr| - \frac{1}{\sqrt{2}} \tan^{-1} \left( \sqrt{\frac{x}{2}} \right) + C$