Answer
$\frac 1 3 (3x^2+6x)^{1/2}+C$
Work Step by Step
$\int \frac {x+1}{\sqrt {3x^2+6x}}dx$
$u=3x^2+6x$
$\frac 1 6 du=x+1$
$\frac1 6 \int \frac{du}{u^{1/2}}$
$\frac 1 6 \times \frac{u^{1/2}}{1/2}$
$\frac 1 3 u^{1/2}+C$
$\frac 1 3 (3x^2+6x)^{1/2}+C$
