부분적분법

To get rid of the square root in the argument of $f''$, we make the change of variables (also called “substitution") $x=t^2,\ \mathrm{d}x=2t\,\mathrm{d}t$.

$$\begin{aligned} \int_0^4 f''\big(\sqrt{x}\big)\,\mathrm{d}x &= 2\int_0^2 tf''(t)\,\mathrm{d}t \end{aligned}$$

Then, to convert $f''$ into $f'$, we integrate by parts with $u=t$, $\mathrm{d}v=f''(t)\,\mathrm{d}t$, $v=f'(t)$.

$$\begin{aligned} \int_0^4 f''\big(\sqrt{x}\big)\,\mathrm{d}x &= 2\bigg\{\Big[\underbrace{tf'(t)}_{uv}\Big]_0^2-\int_0^2\!\!\!\underbrace{ f'(t)\,\mathrm{d}t}_{v\mathrm{d}u}\bigg\} \\ &=2\Big[tf'(t)-f(t)\Big]_0^2 \\ &=2\big[2f'(2)-f(2)+f(0)\big]=2\big[2\times 4-3+1\big]\\ &=12 \end{aligned}$$