Integrating $y'(x)y''(x)$ with respect to $x$

Question

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.

Answer 1

Why the other would be incorrect?

This is because $\int y''(x)~dy \neq y'(x)+C$. However, it is true that: $$\int y''(x)~dx=y'(x)+C$$

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Which is correct?

The second result you obtained is correct (Except you forgot the arbitrary constant). One may easily see this by substituting: $$u=y'(x) \implies du=y''(x)~dx$$ This gives: $$\int u~du=\frac{1}{2}u^2+C=\frac{1}{2}(y'(x))^2+C$$

Answer 2

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

Answer 3

Just to add to projectilemotion's answer, integration by parts may make the calculation easier. \begin{align*} \int y'(x)y''(x)dx = y'(x)^2+c - \int y''(x)y'(x) dx \end{align*}