$$$\frac{4}{15 x^{2} + 27}$$$ 的积分

求$$$\int \frac{4}{15 x^{2} + 27}\, dx$$$。

化简被积函数:

$${\color{red}{\int{\frac{4}{15 x^{2} + 27} d x}}} = {\color{red}{\int{\frac{4}{3 \left(5 x^{2} + 9\right)} d x}}}$$

对 $$$c=\frac{4}{3}$$$ 和 $$$f{\left(x \right)} = \frac{1}{5 x^{2} + 9}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{4}{3 \left(5 x^{2} + 9\right)} d x}}} = {\color{red}{\left(\frac{4 \int{\frac{1}{5 x^{2} + 9} d x}}{3}\right)}}$$

设$$$u=\frac{\sqrt{5}}{3} x$$$。

则$$$du=\left(\frac{\sqrt{5}}{3} x\right)^{\prime }dx = \frac{\sqrt{5}}{3} dx$$$ (步骤见»)，并有$$$dx = \frac{3 \sqrt{5} du}{5}$$$。

所以，

$$\frac{4 {\color{red}{\int{\frac{1}{5 x^{2} + 9} d x}}}}{3} = \frac{4 {\color{red}{\int{\frac{\sqrt{5}}{15 \left(u^{2} + 1\right)} d u}}}}{3}$$

对 $$$c=\frac{\sqrt{5}}{15}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{4 {\color{red}{\int{\frac{\sqrt{5}}{15 \left(u^{2} + 1\right)} d u}}}}{3} = \frac{4 {\color{red}{\left(\frac{\sqrt{5} \int{\frac{1}{u^{2} + 1} d u}}{15}\right)}}}{3}$$

$$$\frac{1}{u^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$\frac{4 \sqrt{5} {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{45} = \frac{4 \sqrt{5} {\color{red}{\operatorname{atan}{\left(u \right)}}}}{45}$$

回忆一下 $$$u=\frac{\sqrt{5}}{3} x$$$:

$$\frac{4 \sqrt{5} \operatorname{atan}{\left({\color{red}{u}} \right)}}{45} = \frac{4 \sqrt{5} \operatorname{atan}{\left({\color{red}{\frac{\sqrt{5}}{3} x}} \right)}}{45}$$

因此，

$$\int{\frac{4}{15 x^{2} + 27} d x} = \frac{4 \sqrt{5} \operatorname{atan}{\left(\frac{\sqrt{5} x}{3} \right)}}{45}$$

加上积分常数：

$$\int{\frac{4}{15 x^{2} + 27} d x} = \frac{4 \sqrt{5} \operatorname{atan}{\left(\frac{\sqrt{5} x}{3} \right)}}{45}+C$$

$$$\int \frac{4}{15 x^{2} + 27}\, dx = \frac{4 \sqrt{5} \operatorname{atan}{\left(\frac{\sqrt{5} x}{3} \right)}}{45} + C$$$A