利用该表（图 8.61），我们选择代换

$$(x + 3) = 4 \cosh u, \quad \text{so that } \frac{dx}{du} = 4 \sinh u \text{ and}$$

$$\sqrt{(x^2 + 6x - 7)} dx = \int \sqrt{(16 \cosh^2 u - 16)} 4 \sinh u du$$

现在 $\cosh^2 u - 1 = \sinh^2 u$ ，积分变为

$$16 \int \sinh^2 u du = 16 \int \frac{1}{2}(\cosh 2u - 1) du = 4 \sinh 2u - 8u + \text{constant}$$

由于 $\cosh u = (x + 3)/4$ 我们推得 $\sinh u = \sqrt{(\cosh^2 u - 1)}$ ；也就是说， $\sinh u =$

$$\sqrt{\left[\left(\frac{x+3}{4}\right)^2 - 1\right]} \text{ and}$$

$$\sinh 2u = 2 \sinh u \cosh u = \frac{2(x+3)}{4} \sqrt{\left[\left(\frac{x+3}{4}\right)^2 - 1\right]}$$

同样 $u = \cosh^{-1}\left(\frac{x+3}{4}\right)$ 因此

$$\int \sqrt{(x^2 + 6x - 7)} dx = (x+3) \sqrt{(x^2 + 6x - 7)} - 8 \cosh^{-1}\left(\frac{x+3}{4}\right) + \text{constant}$$

(b) 使用相同的方法，我们有

$$\int \frac{dx}{\sqrt{(x^2 - 5x + 4)}} = \int \frac{dx}{\sqrt{[(x - \frac{5}{2})^2 - \frac{9}{4}]}}$$

令 $x - \frac{5}{2} = \frac{3}{2} \cosh u$ , 因此 $\frac{dx}{du} = \frac{3}{2} \sinh u$ , 积分变为 $\int \frac{\frac{3}{2} \sinh u}{\sqrt{(\frac{9}{4} \sinh^2 u)}} du$ 由于 $\cosh^2 u - 1 = \sinh^2 u$ 。因此

$$\int \frac{dx}{\sqrt{(x^2 - 5x + 4)}} = \int 1 du = u + c, \text{ where } \cosh u = \frac{2}{3} \left(x - \frac{3}{2}\right)$$

所以

$$\int \frac{dx}{\sqrt{(x^2 - 5x + 4)}} = \cosh^{-1}\left(\frac{2x - 5}{3}\right) + \text{constant}$$

(c) 改写被积函数得

$$\int \frac{dx}{\sqrt{(3x^2 - 6x + 7)}} = \int \frac{dx}{\sqrt{3} \sqrt{[(x - 1)^2 + \frac{4}{3}]}}$$