TITLE: Integrating $y'(x)y''(x)$ with respect to $x$
QUESTION [5 upvotes]: I'm having trouble with integrating $y'y''$, because I can think of two different answers that seem correct to me.
1)  $$y''y' = y'' \frac{dy}{dx}$$
$$y''y'dx = y''dy $$
$$\int y''y'dx = y' + C$$
2) $$y''y' = \frac{d}{dx} (\frac{1}{2} (y')^2)$$
So $$\int y''y'dx = (\frac{1}{2} (y')^2)$$
These answers seem quite different, but I can figure out which is correct (assuming one is) and why the other would be incorrect.

REPLY [5 votes]: You can also go by parts naively taking your first expression: 
$$
\int y'(x)y''(x)dx=(y'(x))^2-\int y'(x)y''(x)dx\implies 2\int y'(x)y''(x)dx=(y'(x))^2\implies \int y'(x)y''(x)dx=\frac{1}{2}(y'(x))^2+c
$$