Integration by Parts? - Variable Manipulation 
$$\int x^3f''(x^2)\,\mathrm{d}x$$ 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. :)

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 $y=x^2$.
