Question

Solve:$\int \frac{1}{\sqrt{ax+b}}dx$

Answer 1

Let $ax+b=t$, then $adx=dt$
$\int \frac{1}{\sqrt{ax+b}}dx$
$=\frac{1}{a}\int \frac{a}{\sqrt{ax+b}}dx$
$=\frac{1}{a}\int \frac{1}{\sqrt{t}}dt$
$=\frac{1}{a}\int t^{-1/2}dt$
$=\frac{1}{a}\frac{t^{1/2}}{1/2}+C$ ($\int x^{n-1}dx=\frac{x^n}{n}+C$)
$=\frac{2}{a}\sqrt{ax+b}+C$