적분 전략 고르기

It's not immediately clear where to start, but a common method we've seen is to use the denominator in a $u$-substitution, especially when square roots are involved.

Let $u = \sqrt{1-x}$, $\mathrm{d}u = -\frac{1}{2\sqrt{1-x}}\mathrm{d}x$. Then $u^2=1-x$, so $x = 1-u^2$.

$$\begin{aligned} \int \sqrt{\frac{x}{1-x}}~\mathrm{d}x&=2\int \frac{\sqrt{x}}{2\sqrt{1-x}}~\mathrm{d}x =- 2\int\sqrt{1-u^2}~\mathrm{d}u \\ \text{Now we're back in familiar territory. Let $u=\sin \theta$, $\mathrm{d}u=\cos \theta~\mathrm{d}\theta$.} \\ &=-2\int\sqrt{1-\sin^2 \theta }\cos\theta~\mathrm{d}\theta\\ &=-2\int\cos^2\theta~\mathrm{d}\theta\\ &=-\int \big(1+\cos(2\theta)\big)~\mathrm{d}\theta\\ &=-\theta - \frac{1}{2}\sin(2\theta)+C\\ &=-\theta - \sin\theta\cos\theta+C\\ &=-\arcsin u - u\sqrt{1-u^2}+C \tag{$*$}\\ \smash{  }\,&=-\arcsin (\sqrt{1-x})-\sqrt{1-x}\sqrt{x}+C \end{aligned}$$

In ($*$), to convert from $\theta$ to $u$, our substitution $u=\sin\theta$ tells us $\theta = \arcsin u$. To find $\cos \theta$, we can either trace our work backwards to see that we already simplified $\sqrt{1-u^2}$ into $\cos\theta$, or we can draw a right triangle with angle $\theta$ and $\sin \theta=u$, then use the Pythagorean theorem to find the length of the adjacent side of the triangle and $\cos\theta$.