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. :)

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. 
