Based on my previous question: (somewhat related to it)
$$\int f''(x^2)~dx$$
How would you go about and find the integral in an abstract sense as you can do the following with derivatives using chain rule:
$$\frac{d}{dx}[f'(x^2)]$$ $$=f''(x^2)2x$$
I kind of call the above abstract as I think of it because we're not given what $f(x)$ actually equals?
Using integration by parts on the integral
$$\int f''(x^2)\,dx=\int \frac1{2x}\frac{d}{dx}\left(f'(x^2)\right)\,dx$$
with $u=1/2x$ and $v=f'(x^2)$ yields
$$\begin{align} \int f''(x^2)\,dx&=\frac{1}{2x}f'(x^2)+\int \frac{1}{2x^2}f'(x^2)\,dx\\\\ &=\frac{1}{2x}f'(x^2)+\int \frac{1}{4x^3}\frac{df(x^2)}{dx}\,dx \tag 1 \end{align}$$
Now, integrating by parts the integral in $(1)$ with $u=1/(4x^3)$ and $v=f(x^2)$ yields
$$\int f''(x^2)\,dx=\frac{1}{2x}f'(x^2)+\frac{1}{4x^3}f(x^2)+\frac34\int \frac{f(x^2)}{x^4}\,dx$$