Let's solve this integral:

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x}{\sqrt[\phantom{2}]{x^2-2}}dx$

$If$ one function looks like it's "inside" another, that inside function $is$

often a good choice for $u.$

$In$ this case, let's choose $u=x^2-2.$ Replace $x^2-2$ with $uin$ the integral:

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x}{\sqrt[\phantom{2}]{x^2-2}}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x}{\sqrt[\phantom{2}]{u}}dx$

Find the derivative $ofu$ :

$\frac{du}{dx}=$

Multiply both sides $bydx$ :

$du=$