삼각치환

This integrand looks very different from those above. But it is only slightly disguised. If we complete the square

$$\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{3-2x-x^2}} = \int \frac{\mathrm{d}x}{\sqrt{4-(x+1)^2}} \end{aligned}$$

and make the substitution $y=x+1$, $\mathrm{d}y=\mathrm{d}x$

$$\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{3-2x-x^2}} = \int \frac{\mathrm{d}x}{\sqrt{4-(x+1)^2}} = \int \frac{\mathrm{d}y}{\sqrt{4-y^2}} \end{aligned}$$

we get a typical trig substitution integral. So, we substitute $y=2\sin\theta$, $\mathrm{d}y=2\cos\theta\,\mathrm{d}\theta$ to get

$$\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{3-2x-x^2}} &= \int \frac{\mathrm{d}y}{\sqrt{4-y^2}} = \int\frac{2\cos\theta\,\mathrm{d}\theta}{\sqrt{4-4\sin^2\theta}} = \int\frac{2\cos\theta\,\mathrm{d}\theta}{\sqrt{4\cos^2\theta}}\\ &=\int\mathrm{d}\theta =\theta +C =\arcsin\frac{y}{2} + C \\ &=\arcsin\frac{x+1}{2} + C \end{aligned}$$

An experienced integrator would probably substitute $x+1 = 2\sin\theta$ directly, without going through $y$.