然后使用线性复合法则， $a = \frac{1}{\sqrt{3}}$ 我们推导出

$$\int_0^1 \frac{dx}{\sqrt{(3-x^2)}} = \left[ \frac{1}{\sqrt{3}} \frac{\sqrt{3}}{1} \sin^{-1}\left(\frac{x}{\sqrt{3}}\right) \right]_0^1 = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$$

(b) 改写被积函数并再次使用线性复合法则：

$$\begin{aligned} \int_0^2 \frac{1}{\sqrt{(3+2x-x^2)}} dx &= \int_0^2 \frac{1}{\sqrt{[4-(x-1)^2]}} dx \ &= \frac{1}{2} \int_0^2 \frac{1}{\sqrt{\left[1 - \left(\frac{x-1}{2}\right)^2\right]}} dx \ &= \left[ \frac{1}{2} \times 2 \sin^{-1}\left(\frac{x-1}{2}\right) \right]_0^2 \ &= \sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{3} \end{aligned}$$

(c) 这里我们回顾 $\int \frac{1}{1+x^2} dx = \tan^{-1}x + c$ 。

改写被积函数，我们得到

$$\int_0^2 \frac{dx}{4+x^2} = \frac{1}{4} \int_0^2 \frac{dx}{1+\left(\frac{x}{2}\right)^2} = \frac{1}{4} \left[ 2 \tan^{-1}\left(\frac{x}{2}\right) \right]_0^2$$

使用线性复合函数法则。因此

$$\int_0^2 \frac{dx}{4+x^2} = \frac{1}{2} \tan^{-1} 1 = \frac{\pi}{8}$$

(d) 改写被积函数，我们得到

$$\begin{aligned} \int_{-5}^5 \frac{1}{x^2 + 10x + 50} dx &= \int_{-5}^5 \frac{1}{(x+5)^2 + 5^2} dx \ &= \frac{1}{25} \int_{-5}^5 \frac{1}{1 + \left(\frac{x}{5} + 1\right)^2} dx \ &= \frac{1}{25} \left[ 5 \tan^{-1}\left(\frac{x}{5} + 1\right) \right]_{-5}^5 \ &= \frac{1}{5} \left[ \tan^{-1} 2 - \tan^{-1} 0 \right] = \frac{1}{5} \tan^{-1} 2 \end{aligned}$$

评注 从这些例子中我们可以推导出另外两个标准积分：

$$\int \frac{dx}{\sqrt{(a^2 - x^2)}} = \sin^{-1} \frac{x}{a} + c$$

$$\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + c$$