$$$\frac{1}{\sqrt{- 6 x^{25} + x^{2}}}$$$ 的积分
您的输入
求$$$\int \frac{1}{\sqrt{- 6 x^{25} + x^{2}}}\, dx$$$。
解答
输入已重写为:$$$\int{\frac{1}{\sqrt{- 6 x^{25} + x^{2}}} d x}=\int{\frac{1}{x \sqrt{1 - 6 x^{23}}} d x}$$$。
设$$$u=x^{\frac{23}{2}}$$$。
则$$$du=\left(x^{\frac{23}{2}}\right)^{\prime }dx = \frac{23 x^{\frac{21}{2}}}{2} dx$$$ (步骤见»),并有$$$x^{\frac{21}{2}} dx = \frac{2 du}{23}$$$。
所以,
$${\color{red}{\int{\frac{1}{x \sqrt{1 - 6 x^{23}}} d x}}} = {\color{red}{\int{\frac{2}{23 u \sqrt{1 - 6 u^{2}}} d u}}}$$
对 $$$c=\frac{2}{23}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u \sqrt{1 - 6 u^{2}}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$${\color{red}{\int{\frac{2}{23 u \sqrt{1 - 6 u^{2}}} d u}}} = {\color{red}{\left(\frac{2 \int{\frac{1}{u \sqrt{1 - 6 u^{2}}} d u}}{23}\right)}}$$
设$$$u=\frac{\sqrt{6} \sin{\left(v \right)}}{6}$$$。
则$$$du=\left(\frac{\sqrt{6} \sin{\left(v \right)}}{6}\right)^{\prime }dv = \frac{\sqrt{6} \cos{\left(v \right)}}{6} dv$$$(步骤见»)。
此外,可得$$$v=\operatorname{asin}{\left(\sqrt{6} u \right)}$$$。
因此,
$$$\frac{1}{ u \sqrt{1 - 6 u ^{2}}} = \frac{\sqrt{6}}{\sqrt{1 - \sin^{2}{\left( v \right)}} \sin{\left( v \right)}}$$$
利用恒等式 $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:
$$$\frac{\sqrt{6}}{\sqrt{1 - \sin^{2}{\left( v \right)}} \sin{\left( v \right)}}=\frac{\sqrt{6}}{\sqrt{\cos^{2}{\left( v \right)}} \sin{\left( v \right)}}$$$
假设$$$\cos{\left( v \right)} \ge 0$$$,我们得到如下结果:
$$$\frac{\sqrt{6}}{\sqrt{\cos^{2}{\left( v \right)}} \sin{\left( v \right)}} = \frac{\sqrt{6}}{\sin{\left( v \right)} \cos{\left( v \right)}}$$$
因此,
$$\frac{2 {\color{red}{\int{\frac{1}{u \sqrt{1 - 6 u^{2}}} d u}}}}{23} = \frac{2 {\color{red}{\int{\frac{1}{\sin{\left(v \right)}} d v}}}}{23}$$
使用二倍角公式 $$$\sin\left( v \right)=2\sin\left(\frac{ v }{2}\right)\cos\left(\frac{ v }{2}\right)$$$ 改写正弦:
$$\frac{2 {\color{red}{\int{\frac{1}{\sin{\left(v \right)}} d v}}}}{23} = \frac{2 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{v}{2} \right)} \cos{\left(\frac{v}{2} \right)}} d v}}}}{23}$$
将分子和分母同时乘以 $$$\sec^2\left(\frac{ v }{2} \right)$$$:
$$\frac{2 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{v}{2} \right)} \cos{\left(\frac{v}{2} \right)}} d v}}}}{23} = \frac{2 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{v}{2} \right)}}{2 \tan{\left(\frac{v}{2} \right)}} d v}}}}{23}$$
设$$$w=\tan{\left(\frac{v}{2} \right)}$$$。
则$$$dw=\left(\tan{\left(\frac{v}{2} \right)}\right)^{\prime }dv = \frac{\sec^{2}{\left(\frac{v}{2} \right)}}{2} dv$$$ (步骤见»),并有$$$\sec^{2}{\left(\frac{v}{2} \right)} dv = 2 dw$$$。
积分变为
$$\frac{2 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{v}{2} \right)}}{2 \tan{\left(\frac{v}{2} \right)}} d v}}}}{23} = \frac{2 {\color{red}{\int{\frac{1}{w} d w}}}}{23}$$
$$$\frac{1}{w}$$$ 的积分为 $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$:
$$\frac{2 {\color{red}{\int{\frac{1}{w} d w}}}}{23} = \frac{2 {\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{23}$$
回忆一下 $$$w=\tan{\left(\frac{v}{2} \right)}$$$:
$$\frac{2 \ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{23} = \frac{2 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{v}{2} \right)}}}}\right| \right)}}{23}$$
回忆一下 $$$v=\operatorname{asin}{\left(\sqrt{6} u \right)}$$$:
$$\frac{2 \ln{\left(\left|{\tan{\left(\frac{{\color{red}{v}}}{2} \right)}}\right| \right)}}{23} = \frac{2 \ln{\left(\left|{\tan{\left(\frac{{\color{red}{\operatorname{asin}{\left(\sqrt{6} u \right)}}}}{2} \right)}}\right| \right)}}{23}$$
回忆一下 $$$u=x^{\frac{23}{2}}$$$:
$$\frac{2 \ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{6} {\color{red}{u}} \right)}}{2} \right)}}\right| \right)}}{23} = \frac{2 \ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{6} {\color{red}{x^{\frac{23}{2}}}} \right)}}{2} \right)}}\right| \right)}}{23}$$
因此,
$$\int{\frac{1}{x \sqrt{1 - 6 x^{23}}} d x} = \frac{2 \ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{6} x^{\frac{23}{2}} \right)}}{2} \right)}}\right| \right)}}{23}$$
加上积分常数:
$$\int{\frac{1}{x \sqrt{1 - 6 x^{23}}} d x} = \frac{2 \ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{6} x^{\frac{23}{2}} \right)}}{2} \right)}}\right| \right)}}{23}+C$$
答案
$$$\int \frac{1}{\sqrt{- 6 x^{25} + x^{2}}}\, dx = \frac{2 \ln\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{6} x^{\frac{23}{2}} \right)}}{2} \right)}}\right|\right)}{23} + C$$$A