$$$\frac{1}{5 - x^{2}}$$$ 的積分
您的輸入
求$$$\int \frac{1}{5 - x^{2}}\, dx$$$。
解答
進行部分分式分解(步驟可見 »):
$${\color{red}{\int{\frac{1}{5 - x^{2}} d x}}} = {\color{red}{\int{\left(\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} - \frac{\sqrt{5}}{10 \left(x - \sqrt{5}\right)}\right)d x}}}$$
逐項積分:
$${\color{red}{\int{\left(\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} - \frac{\sqrt{5}}{10 \left(x - \sqrt{5}\right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{\sqrt{5}}{10 \left(x - \sqrt{5}\right)} d x} + \int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x}\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{\sqrt{5}}{10}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x - \sqrt{5}}$$$:
$$\int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x} - {\color{red}{\int{\frac{\sqrt{5}}{10 \left(x - \sqrt{5}\right)} d x}}} = \int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x} - {\color{red}{\left(\frac{\sqrt{5} \int{\frac{1}{x - \sqrt{5}} d x}}{10}\right)}}$$
令 $$$u=x - \sqrt{5}$$$。
則 $$$du=\left(x - \sqrt{5}\right)^{\prime }dx = 1 dx$$$ (步驟見»),並可得 $$$dx = du$$$。
該積分變為
$$\int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x} - \frac{\sqrt{5} {\color{red}{\int{\frac{1}{x - \sqrt{5}} d x}}}}{10} = \int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x} - \frac{\sqrt{5} {\color{red}{\int{\frac{1}{u} d u}}}}{10}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x} - \frac{\sqrt{5} {\color{red}{\int{\frac{1}{u} d u}}}}{10} = \int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x} - \frac{\sqrt{5} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{10}$$
回顧一下 $$$u=x - \sqrt{5}$$$:
$$- \frac{\sqrt{5} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{10} + \int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x} = - \frac{\sqrt{5} \ln{\left(\left|{{\color{red}{\left(x - \sqrt{5}\right)}}}\right| \right)}}{10} + \int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$,使用 $$$c=\frac{\sqrt{5}}{10}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x + \sqrt{5}}$$$:
$$- \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + {\color{red}{\int{\frac{\sqrt{5}}{10 \left(x + \sqrt{5}\right)} d x}}} = - \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + {\color{red}{\left(\frac{\sqrt{5} \int{\frac{1}{x + \sqrt{5}} d x}}{10}\right)}}$$
令 $$$u=x + \sqrt{5}$$$。
則 $$$du=\left(x + \sqrt{5}\right)^{\prime }dx = 1 dx$$$ (步驟見»),並可得 $$$dx = du$$$。
該積分變為
$$- \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + \frac{\sqrt{5} {\color{red}{\int{\frac{1}{x + \sqrt{5}} d x}}}}{10} = - \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + \frac{\sqrt{5} {\color{red}{\int{\frac{1}{u} d u}}}}{10}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + \frac{\sqrt{5} {\color{red}{\int{\frac{1}{u} d u}}}}{10} = - \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + \frac{\sqrt{5} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{10}$$
回顧一下 $$$u=x + \sqrt{5}$$$:
$$- \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + \frac{\sqrt{5} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{10} = - \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + \frac{\sqrt{5} \ln{\left(\left|{{\color{red}{\left(x + \sqrt{5}\right)}}}\right| \right)}}{10}$$
因此,
$$\int{\frac{1}{5 - x^{2}} d x} = - \frac{\sqrt{5} \ln{\left(\left|{x - \sqrt{5}}\right| \right)}}{10} + \frac{\sqrt{5} \ln{\left(\left|{x + \sqrt{5}}\right| \right)}}{10}$$
化簡:
$$\int{\frac{1}{5 - x^{2}} d x} = \frac{\sqrt{5} \left(- \ln{\left(\left|{x - \sqrt{5}}\right| \right)} + \ln{\left(\left|{x + \sqrt{5}}\right| \right)}\right)}{10}$$
加上積分常數:
$$\int{\frac{1}{5 - x^{2}} d x} = \frac{\sqrt{5} \left(- \ln{\left(\left|{x - \sqrt{5}}\right| \right)} + \ln{\left(\left|{x + \sqrt{5}}\right| \right)}\right)}{10}+C$$
答案
$$$\int \frac{1}{5 - x^{2}}\, dx = \frac{\sqrt{5} \left(- \ln\left(\left|{x - \sqrt{5}}\right|\right) + \ln\left(\left|{x + \sqrt{5}}\right|\right)\right)}{10} + C$$$A