$\displaystyle \int \frac{u}{\sqrt{u^2 +16}} \ du$
Thanks in advance.
$\displaystyle \int \frac{u}{\sqrt{u^2 + 16}} \ du$
$\displaystyle w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}$
$\displaystyle \int \frac{u}{\sqrt w} . \frac{dw}{2u}$
$\displaystyle \int \frac{1}{\sqrt w} . \frac{dw}{2}$
$\displaystyle \frac{1}{2} \int \frac{1}{\sqrt w} \ dw$
$\displaystyle \frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right] $
$\displaystyle w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}$