一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
$$
\int \sqrt{\frac{3-2 x}{3+2 x}} \mathrm{d} x = ?
$$
难度评级:
二、解析 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
对于含有根式的题目,一个主要的解题思路就是想办法去除根号,关于根式的一些性质可以查看《根式的常用性质》这篇文章。
解法 1:分子有理化
$$
\int \sqrt{\frac{3-2 x}{3+2 x}} \mathrm{d} x =
$$
$$
\int \sqrt{\frac{(3 – 2 x)(3 – 2 x)}{(3 + 2 x)(3 – 2 x)}} \mathrm{d} x =
$$
$$
\int \frac{3-2 x}{\sqrt{9 – 4 x^{2}}} \mathrm{d} x =
$$
$$
\int \frac{3}{\sqrt{9 – 4 x^{2}}} \mathrm{d} x + \int \frac{-2 x}{\sqrt{9 – 4 x^{2}}} \mathrm{d} x =
$$
$$
\int \frac{3}{\sqrt{9[1 – (\frac{2}{3} x)^{2}]}} \mathrm{d} x + \textcolor{red}{\frac{1}{4} } \int \frac{(9 – 4 x^{2})^{\prime} }{\sqrt{9 – 4 x^{2}}} \mathrm{d} x =
$$
注意:$(-4x^{2})^{\prime}$ $=$ $-8x$.
Next
$$
\int \frac{3}{3 \sqrt{1 – (\frac{2}{3} x)^{2}}} \mathrm{d} x + \frac{1}{4} \int \frac{\mathrm{d} (9 – 4 x^{2}) }{\sqrt{9 – 4 x^{2}}} =
$$
$$
\textcolor{red}{\frac{3}{2} } \int \frac{1}{\sqrt{1 – (\frac{2}{3} x)^{2}}} \mathrm{d} \Big(\frac{2}{3} x \Big) + \frac{1}{4} \int (9 – 4 x^{2})^{ \textcolor{red}{\frac{-1}{2}}} \mathrm{d} (9 – 4 x^{2}) =
$$
$$
\frac{3}{2} \arcsin (\frac{2}{3} x) + \frac{1}{4} \cdot 2 \cdot \sqrt{9 – 4x^{2}} + C =
$$
进行积分运算时,可以通过对积分所得的结果进行求导运算的方式进行检验:$(2 \cdot \sqrt{9 – 4x^{2}})^{\prime}$ $=$ $2 \cdot \frac{1}{2} (9 – 4x^{2})^{\frac{-1}{2}}$
Next
$$
\frac{3}{2} \arcsin (\frac{2}{3} x) + \frac{1}{2} \cdot \sqrt{9 – 4x^{2}} + C.
$$
其中,$C$ 为任意常数。
解法 2:整体代换
$$
\int \sqrt{\frac{3-2 x}{3+2 x}} \mathrm{d} x \Rightarrow
$$
令 $t = \sqrt{\frac{3-2 x}{3+2 x}}$, 则:
$$
t^{2} = \frac{3-2 x}{3+2 x} \Rightarrow
$$
$$
t^{2} (3 + 2x) = 3 – 2x \Rightarrow
$$
$$
3t^{2} + 2t^{2} x = 3 – 2x \Rightarrow
$$
$$
2t^{2} x + 2x = 3 – 3t^{2} \Rightarrow
$$
$$
x(2t^{2} + 2) = 3 – 3t^{2} \Rightarrow
$$
$$
x = \frac{3 – 3t^{2}}{2 + 2t^{2}} \Rightarrow
$$
$$
x = \frac{3}{2} \cdot \frac{1-t^{2}}{1 + t^{2}}.
$$
Next
于是:
$$
\mathrm{d} x = \frac{3}{2} \Big( \frac{1-t^{2}}{1 + t^{2}} \Big)^{\prime} \Rightarrow
$$
$$
\mathrm{d} x = \frac{3}{2} \cdot \frac{-2t(1 + t^{2}) – 2t(1-t^{2})}{ (1 + t^{2})^{2} } \textcolor{red}{ \mathrm{d} t } \Rightarrow
$$
$$
\mathrm{d} x = \frac{3}{2} \cdot \frac{-2t – 2t^{3} – 2t + 2t^{3}}{ (1 + t^{2})^{2} } \textcolor{red}{ \mathrm{d} t } \Rightarrow
$$
$$
\mathrm{d} x = \frac{3}{2} \cdot \frac{-4t}{ (1 + t^{2})^{2} } \textcolor{red}{ \mathrm{d} t } \Rightarrow
$$
$$
\mathrm{d} x = \frac{-6t}{ (1 + t^{2})^{2} } \textcolor{red}{ \mathrm{d} t }.
$$
Next
进而:
$$
\int \sqrt{\frac{3-2 x}{3+2 x}} \mathrm{d} x =
$$
$$
\int t \cdot \frac{-6t}{ (1 + t^{2})^{2} } \textcolor{red}{ \mathrm{d} t } =
$$
$$
\int \frac{-6t^{2}}{ (1 + t^{2})^{2} } \mathrm{d} t.
$$
Next
又:
$$
\Big( \frac{1}{1 + t^{2}} \Big)^{\prime} = \frac{ \textcolor{red}{-} 2t }{(1 + t^{2})^{2}}
$$
Next
于是:
$$
\int \frac{-6t^{2}}{ (1 + t^{2})^{2} } \mathrm{d} t =
$$
$$
\int 3t \cdot \mathrm{d} \Big( \frac{1}{1 + t^{2}} \Big) =
$$
$$
\frac{3t}{1 + t^{2}} – 3 \int \frac{1}{1 + t^{2}} \mathrm{d} t =
$$
$$
\frac{3t}{1 + t^{2}} – 3 \arctan t + C \Rightarrow
$$
Next
令 $t = \sqrt{\frac{3-2 x}{3+2 x}}$ $\Rightarrow$
$$
\frac{3\sqrt{\frac{3-2 x}{3+2 x}}}{1 + \frac{3-2 x}{3+2 x}} – 3 \arctan \sqrt{\frac{3-2 x}{3+2 x}} + C \Rightarrow
$$
$$
\frac{3\sqrt{\frac{3-2 x}{3+2 x}}}{ \frac{6}{3+2 x}} – 3 \arctan \sqrt{\frac{3-2 x}{3+2 x}} + C \Rightarrow
$$
$$
\sqrt{\frac{3-2 x}{3+2 x}} \cdot \frac{3+2 x}{2} – 3 \arctan \sqrt{\frac{3-2 x}{3+2 x}} + C \Rightarrow
$$
$$
\sqrt{\frac{(3-2 x) (3+2 x)^{2}}{4(3+2 x)}} – 3 \arctan \sqrt{\frac{3-2 x}{3+2 x}} + C \Rightarrow
$$
$$
\sqrt{\frac{(3-2 x) (3+2 x)}{4}} – 3 \arctan \sqrt{\frac{3-2 x}{3+2 x}} + C \Rightarrow
$$
$$
\sqrt{\frac{ 9 – 4x^{2} }{4}} – 3 \arctan \sqrt{\frac{3-2 x}{3+2 x}} + C \Rightarrow
$$
$$
\frac{1}{2}\sqrt{ 9 – 4x^{2}} – 3 \arctan \sqrt{\frac{3-2 x}{3+2 x}} + C.
$$
其中,$C$ 为任意常数。
注意:使用上述两种方法计算所得的结果是不一样的,但都是正确的。
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