For example, let's solve ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x{e}^{x^2}dx$ using this technique.

$u(x)=x^2(the$ inside thing!$)$

$dx=\frac{du}{u^{\text{'}}}=\frac{du}{2x}$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x{e}^{x^{u^{\text{'}}}}dx={\displaystyle \underset{\phantom{}}{\overset{2x}{}}}x{e}^u\frac{du}{2x},$ after substituting $u$ and $dx$ away

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x{e}^u\frac{du}{2x}={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{2}{e}^{u}du$ after simplifying. I All the $x^{\text{'}}s$ will cancel

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{2}{e}^{u}du=\frac{e^u}{2}+C=\frac{e^{x^2}}{2}+C$ after integrating.

Evaluate the indefinite integral using the above technique.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{cosx}{(7sin(x)+21{)}^3}dx$

$u(x)=$

$u^{\text{'}}(x)=,(take$ the derivative$)$

a fraction involving $du$ and $\{$:$u^{\text{'}})$

After substituting $u$ and $dx$ into the integral, the integral $is$ just $in$ terms $ofu$ and $du$ :

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$

Concluding,

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{cosx}{(7sin(x)+21{)}^3}dx=$