求 $\int \frac{x+2}{x^2+2x-3} \,\mathrm{d}x$。 求 $\int 2x\sin x \,\mathrm{d}x$。 利用换元法 $x = 3\sin\theta$，求 $\int_0^{\frac{3}{2}} \sqrt{9-x^2} \,\mathrm{d}x$ 的精确值。

1. Express the integrand in partial fractions.

The denominator factorises as $x^2+2x-3 = (x-1)(x+3)$.
Let $\frac{x+2}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3}$.
Multiply throughout by $(x-1)(x+3)$:
$x+2 = A(x+3) + B(x-1)$

2. Find the constants $A$ and $B$.

Substitute $x = 1$:
$1 + 2 = A(1 + 3) \implies 3 = 4A \implies A = \frac{3}{4}$

Substitute $x = -3$:
$-3 + 2 = B(-3 - 1) \implies -1 = -4B \implies B = \frac{1}{4}$

Thus, $\frac{x+2}{x^2+2x-3} = \frac{3}{4(x-1)} + \frac{1}{4(x+3)}$.

3. Integrate the resulting partial fractions.

$\int \frac{x+2}{x^2+2x-3} \,\mathrm{d}x = \int \left( \frac{3}{4(x-1)} + \frac{1}{4(x+3)} \right) \,\mathrm{d}x$
$= \frac{3}{4}\ln|x-1| + \frac{1}{4}\ln|x+3| + C$, where $C$ is an arbitrary constant.

1. Set up integration by parts by identifying $u$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$.

Let $u = 2x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2$
Let $\frac{\mathrm{d}v}{\mathrm{d}x} = \sin x \implies v = -\cos x$

2. Apply the integration by parts formula and evaluate the resulting integral.

$\begin{aligned} \int 2x\sin x \,\mathrm{d}x &= uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \,\mathrm{d}x \\ &= 2x(-\cos x) - \int (-\cos x)(2) \,\mathrm{d}x \\ &= -2x\cos x + 2\int \cos x \,\mathrm{d}x \\ &= -2x\cos x + 2\sin x + C \end{aligned}$

1. Find the differential $\mathrm{d}x$ in terms of $\theta$.

$x = 3\sin\theta \implies \frac{\mathrm{d}x}{\mathrm{d}\theta} = 3\cos\theta \implies \mathrm{d}x = 3\cos\theta \,\mathrm{d}\theta$

2. Find the new limits of integration for $\theta$.

When $x = 0$:
$3\sin\theta = 0 \implies \sin\theta = 0 \implies \theta = 0$

When $x = \frac{3}{2}$:
$3\sin\theta = \frac{3}{2} \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$

3. Substitute $x$, $\mathrm{d}x$ and the limits into the integral and simplify the integrand.

$\int_0^{\frac{3}{2}} \sqrt{9-x^2} \,\mathrm{d}x = \int_0^{\frac{\pi}{6}} \sqrt{9-(3\sin\theta)^2} (3\cos\theta) \,\mathrm{d}\theta$
$= \int_0^{\frac{\pi}{6}} \sqrt{9-9\sin^2\theta} (3\cos\theta) \,\mathrm{d}\theta$
$= \int_0^{\frac{\pi}{6}} \sqrt{9(1-\sin^2\theta)} (3\cos\theta) \,\mathrm{d}\theta$
$= \int_0^{\frac{\pi}{6}} (3\cos\theta)(3\cos\theta) \,\mathrm{d}\theta$
$= \int_0^{\frac{\pi}{6}} 9\cos^2\theta \,\mathrm{d}\theta$

4. Use the double angle formula to prepare the integrand for evaluation.

$= 9 \int_0^{\frac{\pi}{6}} \left( \frac{1+\cos 2\theta}{2} \right) \,\mathrm{d}\theta$
$= \frac{9}{2} \int_0^{\frac{\pi}{6}} (1+\cos 2\theta) \,\mathrm{d}\theta$

5. Integrate and evaluate at the limits to find the exact value.

$= \frac{9}{2} \left[ \theta + \frac{1}{2}\sin 2\theta \right]_0^{\frac{\pi}{6}}$
$= \frac{9}{2} \left[ \left( \frac{\pi}{6} + \frac{1}{2}\sin \frac{\pi}{3} \right) - (0 + 0) \right]$
$= \frac{9}{2} \left( \frac{\pi}{6} + \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) \right)$
$= \frac{9}{2} \left( \frac{\pi}{6} + \frac{\sqrt{3}}{4} \right)$
$= \frac{3\pi}{4} + \frac{9\sqrt{3}}{8}$