Question

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

Answer 1

Let ax + b = t

$\Rightarrow$ adx = dt

∴ dx = $\frac{1}{a}dt$

$\Rightarrow \int (ax+b)^{\frac{1}{2}}dx=\frac{1}{a}\int t^{\frac{1}{2}}dt$

= $\frac{1}{a}\bigg(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\bigg)+C$

= $\frac{2}{3a}(ax+b)^{\frac{3}{2}}+C$