1
$\begingroup$

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.

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

setting $t=3x^2$ we get $dx=\frac{1}{6x}dt$ thus we have $\frac{1}{6}\int f'(t)dt$

$\endgroup$
0
$\begingroup$

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 $xf(3x^2)=\dfrac{1}{6} (6xf(3x^2))$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.