Since $\cosh u = \frac{x+2}{\sqrt{7}}$ , $\sinh u = \sqrt{\left[\left(\frac{x+2}{\sqrt{7}}\right)^2 - 1\right]}$ 我们得到

$$\int x\sqrt{x^2 + 4x - 3} dx = \frac{7\sqrt{7}}{3} \frac{1}{7\sqrt{7}} [(x+2)^2 - 7]^{3/2} - (x+2)[(x+2)^2 - 7]^{1/2} + 7 \cosh^{-1}(\frac{x+2}{\sqrt{7}}) + \text{constant}$$

$$= \frac{1}{3}(x^2 + 4x - 3)^{3/2} - (x+2)(x^2 + 4x - 3)^{1/2} + 7 \cosh^{-1}(\frac{x+2}{\sqrt{7}}) + \text{constant}$$

(f) Here $3 + 2x - 2x^2 = \frac{7}{2} - 2(x - \frac{1}{2})^2 = 2[\frac{7}{4} - (x - \frac{1}{2})^2]$ , 所以我们选取 $x - \frac{1}{2} = \frac{\sqrt{7}}{2} \sin u$ 且积分变为 $\int \sqrt{3 + 2x - 2x^2} dx = \sqrt{2} [\frac{7}{4} \cos^2 u du$ since $1 - \sin^2 u = \cos^2 u$ .

Now $\int \frac{7}{4} \cos^2 u du = \int \frac{7}{8} (\cos 2u + 1) du = \frac{7}{16} \sin 2u + \frac{7}{8} u + \text{constant}$

$$= \frac{7}{8} \sin u \cos u + \frac{7}{8} u + \text{constant}$$

回代后我们得到

$$\int \sqrt{3 + 2x - 2x^2} dx = \frac{7\sqrt{2}}{8} \sin^{-1}(\frac{2x-1}{\sqrt{7}}) + \frac{7\sqrt{2}}{8} \frac{2x-1}{\sqrt{7}} \sqrt{1 - (\frac{2x-1}{\sqrt{7}})^2} + \text{constant}$$

$$= \frac{7\sqrt{2}}{8} \sin(\frac{2x-1}{\sqrt{7}}) + \frac{1}{4}(2x-1)\sqrt{3 + 2x - 2x^2} + \text{constant}$$

Comment 这些例子说明了这类积分的复杂性，这也正是推动开发 MAPLE 等计算机软件包的动因。.

8.8.14 Exercises

尽可能使用 MATLAB 检验你的答案.

121 利用给定的代换求下列函数的积分:

(a) $x^3\sqrt{1+x^2}$ , with $t = \sqrt{1+x^2}$

(b) $\frac{3}{x\sqrt{x^2+9}}$ with $t = \frac{1}{x}$

(c) $\frac{1}{3+\sqrt{x}}$ with $t = \sqrt{x}$

122 用适当的换元法求下列函数的积分:

(a) $\frac{1}{1+\sqrt{1+x}}$ (b) $\sin^2 x \cos^3 x$ (c) $\sin \sqrt{x}$

123 证明 $t = \tan \frac{1}{2}x$ implies

$$\sin x = \frac{2t}{1+t^2}$$

$$\cos x = \frac{1-t^2}{1+t^2}$$

and

$$dx = \frac{2}{1+t^2} dt$$

因此积分

(a) $\cos e x$

(b) $\sec x$

(c) $\frac{1}{3+4\sin x}$

(d) $\frac{1}{5\sin x + 12\cos x}$