삼각치환

We use the substitution $x=\tan u$, $\mathrm{d}x=\sec^2 u~\mathrm{d}u$. Note $\tan 0=0$ and $\tan \frac{\pi}{4} =1$.

$$\begin{aligned} \displaystyle\int_0^1\dfrac{x^2}{\sqrt{x^2+1}^3}\mathrm{d}x&= \int_0^{\pi/4}\dfrac{\tan^2 u}{\sqrt{\tan^2 u +1}^3}\sec^2u~\mathrm{d}u\\ &= \int_0^{\pi/4}\dfrac{\tan^2 u}{\sqrt{\sec^2u}^3}\sec^2u~\mathrm{d}u\\ &= \int_0^{\pi/4}\frac{\tan^2 u}{\sec u}~\mathrm{d}u\\&= \int_0^{\pi/4}\frac{\sec^2 u-1}{\sec u}~\mathrm{d}u\\&= \int_0^{\pi/4}\big({\sec u}-\cos u\big)~\mathrm{d}u\\ &=\Big[\log\left|\sec u + \tan u \right| - \sin u\Big]_{0}^{\pi/4} \\ &=\left(\log\left|\sqrt{2} + 1 \right| -\frac{1}{\sqrt{2}}\right) - \left(\log|1+0|-0\right)\\ &=\log(1+\sqrt{2})-\frac{1}{\sqrt{2}} \end{aligned}$$