Answer
$\int 2x\sqrt {3x^2-1}dx=\frac{2}{9}(3x^2-1)^{\frac{3}{2}}+C$
Work Step by Step
Substitution:
$u=3x^2-1$
$\frac{du}{dx}=6x$
$dx=\frac{1}{6x}du$
$\int 2x\sqrt {3x^2-1}dx=\int2x\sqrt u\frac{1}{6x}du=\int\frac{1}{3}u^{\frac{1}{2}}du=\frac{1}{3}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}(3x^2-1)^{\frac{3}{2}}+C$
