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

Let $ax+b = t$.
Differentiating both sides with respect to $x$,we get $a \, dx = dt$,which implies $dx = \frac{1}{a} \, dt$.
Substituting these into the integral:
$\int \sqrt{ax+b} \, dx = \int t^{1/2} \cdot \frac{1}{a} \, dt$
$= \frac{1}{a} \int t^{1/2} \, dt$
$= \frac{1}{a} \left( \frac{t^{1/2+1}}{1/2+1} \right) + C$
$= \frac{1}{a} \left( \frac{t^{3/2}}{3/2} \right) + C$
$= \frac{2}{3a} t^{3/2} + C$
Substituting $t = ax+b$ back,we get:
$= \frac{2}{3a} (ax+b)^{3/2} + C$,where $C$ is an arbitrary constant.