Question

Integrate the function : $\sqrt{ax+b}$

Answer 1

Put $ax+b=t$
$\Rightarrow adx=dt$
$\Rightarrow dx=\frac{1}{a}dt$
$\Rightarrow \int (ax+b{)}^{\frac{1}{2}}dx=\frac{1}{a}\int t^{\frac{1}{2}}dt$
$=\frac{1}{a}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
$=\frac{2}{3a}(ax+b{)}^{\frac{3}{2}}+C$