Integration by Parts? - Variable Manipulation

Question

$$\int x^3f''(x^2)\,\mathrm{d}x$$ Solve using Integration by Parts. \begin{align} u&=x^3\qquad\mathrm{d}v=f''(x^2) \\ \mathrm{d}u&=3x^2\qquad v=f'(x^2) \\ &=x^3f'(x)-\int f'(x^2)3x^2 \\ u&=3x^2\quad\mathrm{d}v=f'(x^2) \\ \mathrm{d}u&=6x \qquad v=f(x^2) \\ &=x^3f'(x^2)-[3x^2f(x^2)-\int f(x^2)6x] \end{align}

No clue what do from here as the correct answer is:

$$\frac{1}{2}(x^2f'(x^2)-f(x^2))+C$$

Can you guys think of anything? I appreciate the help. :)

Answer 1

Hint: The derivative of $f'(x^2)$ is $f''(x^2) \cdot 2x$ using the chain rule. (it isn't just $f'(x^2)$ as you have it now.) So at your first step you need to use up one of the $x$'s from the $x^3$ factor, and adjust for the extra power of 2. With this idea applied twice it may work better.

Answer 2

I suggest you start with a small change of variable (just to make life easier) since $$I=\int x^3f''(x^2)~dx=\frac 12\int x^2 f''(x) 2x dx$$ S0, let $x^2=y$, $2x\,dx=dy$ which make $$I=\frac 12\int y\,f''(y) \, dy$$ Now, one integration by parts $$u=y\implies u'=dy$$ $$dv=f''(y)dy\implies v=f'(y)$$ So $$2I=u\, v-\int v\,u'=y \,f'(y)-\int f'(y)\,dy=y \,f'(y)-f(y)+C$$ Go back to $x$ using $y=x^2$.