$$$\frac{1}{\sqrt{y \left(y - 1\right)}}$$$ 的积分

该计算器将求出$$$\frac{1}{\sqrt{y \left(y - 1\right)}}$$$的积分/原函数,并显示步骤。

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您的输入

$$$\int \frac{1}{\sqrt{y \left(y - 1\right)}}\, dy$$$

解答

输入已重写为:$$$\int{\frac{1}{\sqrt{y \left(y - 1\right)}} d y}=\int{\frac{1}{\sqrt{y^{2} - y}} d y}$$$

配平方(步骤见»):$$$y^{2} - y = \left(y - \frac{1}{2}\right)^{2} - \frac{1}{4}$$$:

$${\color{red}{\int{\frac{1}{\sqrt{y^{2} - y}} d y}}} = {\color{red}{\int{\frac{1}{\sqrt{\left(y - \frac{1}{2}\right)^{2} - \frac{1}{4}}} d y}}}$$

$$$u=y - \frac{1}{2}$$$

$$$du=\left(y - \frac{1}{2}\right)^{\prime }dy = 1 dy$$$ (步骤见»),并有$$$dy = du$$$

因此,

$${\color{red}{\int{\frac{1}{\sqrt{\left(y - \frac{1}{2}\right)^{2} - \frac{1}{4}}} d y}}} = {\color{red}{\int{\frac{1}{\sqrt{u^{2} - \frac{1}{4}}} d u}}}$$

$$$u=\frac{\cosh{\left(v \right)}}{2}$$$

$$$du=\left(\frac{\cosh{\left(v \right)}}{2}\right)^{\prime }dv = \frac{\sinh{\left(v \right)}}{2} dv$$$(步骤见»)。

此外,可得$$$v=\operatorname{acosh}{\left(2 u \right)}$$$

因此,

$$$\frac{1}{\sqrt{ u ^{2} - \frac{1}{4}}} = \frac{1}{\sqrt{\frac{\cosh^{2}{\left( v \right)}}{4} - \frac{1}{4}}}$$$

利用恒等式 $$$\cosh^{2}{\left( v \right)} - 1 = \sinh^{2}{\left( v \right)}$$$

$$$\frac{1}{\sqrt{\frac{\cosh^{2}{\left( v \right)}}{4} - \frac{1}{4}}}=\frac{2}{\sqrt{\cosh^{2}{\left( v \right)} - 1}}=\frac{2}{\sqrt{\sinh^{2}{\left( v \right)}}}$$$

假设$$$\sinh{\left( v \right)} \ge 0$$$,我们得到如下结果:

$$$\frac{2}{\sqrt{\sinh^{2}{\left( v \right)}}} = \frac{2}{\sinh{\left( v \right)}}$$$

积分变为

$${\color{red}{\int{\frac{1}{\sqrt{u^{2} - \frac{1}{4}}} d u}}} = {\color{red}{\int{1 d v}}}$$

应用常数法则 $$$\int c\, dv = c v$$$,使用 $$$c=1$$$

$${\color{red}{\int{1 d v}}} = {\color{red}{v}}$$

回忆一下 $$$v=\operatorname{acosh}{\left(2 u \right)}$$$:

$${\color{red}{v}} = {\color{red}{\operatorname{acosh}{\left(2 u \right)}}}$$

回忆一下 $$$u=y - \frac{1}{2}$$$:

$$\operatorname{acosh}{\left(2 {\color{red}{u}} \right)} = \operatorname{acosh}{\left(2 {\color{red}{\left(y - \frac{1}{2}\right)}} \right)}$$

因此,

$$\int{\frac{1}{\sqrt{y^{2} - y}} d y} = \operatorname{acosh}{\left(2 y - 1 \right)}$$

加上积分常数:

$$\int{\frac{1}{\sqrt{y^{2} - y}} d y} = \operatorname{acosh}{\left(2 y - 1 \right)}+C$$

答案

$$$\int \frac{1}{\sqrt{y \left(y - 1\right)}}\, dy = \operatorname{acosh}{\left(2 y - 1 \right)} + C$$$A


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