0

I stuck to find the integral $\int f(x)f'''(x)\ dx$.

I found the rule $\int f(x)f'(x)\ dx=\frac{1}{2}f^2(x)+c$. How can I proceed?

IV_
  • 6,964
angela93
  • 129

1 Answers1

2

$$\int f(x)f'''(x)dx=f(x)f''(x)-\int f'(x)f''(x)dx$$ Now: $$\int f'(x)f''(x)dx=f'(x)^2-\int f''(x)f'(x)dx$$ So you can show: $$\int f'(x)f''(x)dx=\frac12f'(x)^2$$ And so: $$\int f'(x)f'''(x)=f(x)f''(x)-\frac12f'(x)^2+C$$

Henry Lee
  • 12,215
  • 1
    The second integration by parts is useless. $f'f''$ immediately integrates as $f'^2/2$. –  Feb 20 '20 at 21:45