# Help Evaluating $\int\left(xf'(3x^2)\right)dx$

Does anyone know how to evaluate the following integral?

$\int\left(xf'(3x^2)\right)dx$

The answer is $\frac16 f(3x^2) + C$ , but I want to see a step by step solution if possible.

• Multiply the given integral by $6$ inside and by $\frac 1 6$ outside. Then recognize something that looks like $\int g'\cdot(h'\circ g)$. – Git Gud Oct 26 '14 at 15:40

setting $t=3x^2$ we get $dx=\frac{1}{6x}dt$ thus we have $\frac{1}{6}\int f'(t)dt$
Think about formula $(u(v(x)))'=v'(x)u(v(x))$
if the formula can be used here then $v(x)=3x^2$ so compute the derivative:
$v'(x)=6x$ and the result follows since $xf(3x^2)=\dfrac{1}{6} (6xf(3x^2))$