Question

Solve:
$\int \sqrt{ax+b}dx$

Answer 1

$\int \sqrt{ax+b}.dx$

Let $ax+b=t$

diff w.r.t 'x'

$a\left(1\right)+0=\frac{dt}{dx}$

$dx=\frac{dt}{a}$

now

$\int \sqrt{t}.\frac{dt}{a}$

= $\frac{1}{a}\int \sqrt{t}.dt$

= $\frac{1}{a}\int t^{1/2}.dt$

= $\frac{1}{a}[\frac{t^{3/2}}{3/2}+C]$

= $\frac{1}{a}.[\frac{2}{3}{t}^{3/2}+C]$

= $\frac{2}{3a}{t}^{3/2}+C$

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