Solution assignment 10 Integration by substitution

Solve:

$\displaystyle\int{\sqrt{3x+1}}dx$

We rewrite the integrand:

$\sqrt{3x+1}=(3x+1)^{\frac{1}{2}}$

and the primitive is:

$\displaystyle\frac{1}{\displaystyle\frac{1}{2}+1}(3x+1)^{\frac{1}{2}+1}.\displaystyle\frac{1}{3}=\displaystyle\frac{2}{9}(3x+1)^{\frac{3}{2}}$

and thus the original integral becomes:

$\displaystyle\int{\sqrt{3x+1}}dx=\displaystyle\frac{2}{9}(3x+1)\sqrt{3x+1}+C$

We can verify the result by differentiation. The chain rule has to be applied.