$$$\frac{x^{3}}{x^{2} - 9}$$$ 的積分
您的輸入
求$$$\int \frac{x^{3}}{x^{2} - 9}\, dx$$$。
解答
由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:
$${\color{red}{\int{\frac{x^{3}}{x^{2} - 9} d x}}} = {\color{red}{\int{\left(x + \frac{9 x}{x^{2} - 9}\right)d x}}}$$
逐項積分：
$${\color{red}{\int{\left(x + \frac{9 x}{x^{2} - 9}\right)d x}}} = {\color{red}{\left(\int{x d x} + \int{\frac{9 x}{x^{2} - 9} d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：
$$\int{\frac{9 x}{x^{2} - 9} d x} + {\color{red}{\int{x d x}}}=\int{\frac{9 x}{x^{2} - 9} d x} + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\int{\frac{9 x}{x^{2} - 9} d x} + {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
令 $$$u=x^{2} - 9$$$。
則 $$$du=\left(x^{2} - 9\right)^{\prime }dx = 2 x dx$$$ (步驟見»)，並可得 $$$x dx = \frac{du}{2}$$$。
該積分變為
$$\frac{x^{2}}{2} + {\color{red}{\int{\frac{9 x}{x^{2} - 9} d x}}} = \frac{x^{2}}{2} + {\color{red}{\int{\frac{9}{2 u} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{9}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：
$$\frac{x^{2}}{2} + {\color{red}{\int{\frac{9}{2 u} d u}}} = \frac{x^{2}}{2} + {\color{red}{\left(\frac{9 \int{\frac{1}{u} d u}}{2}\right)}}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：
$$\frac{x^{2}}{2} + \frac{9 {\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{x^{2}}{2} + \frac{9 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
回顧一下 $$$u=x^{2} - 9$$$：
$$\frac{x^{2}}{2} + \frac{9 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{x^{2}}{2} + \frac{9 \ln{\left(\left|{{\color{red}{\left(x^{2} - 9\right)}}}\right| \right)}}{2}$$
因此，
$$\int{\frac{x^{3}}{x^{2} - 9} d x} = \frac{x^{2}}{2} + \frac{9 \ln{\left(\left|{x^{2} - 9}\right| \right)}}{2}$$
加上積分常數：
$$\int{\frac{x^{3}}{x^{2} - 9} d x} = \frac{x^{2}}{2} + \frac{9 \ln{\left(\left|{x^{2} - 9}\right| \right)}}{2}+C$$
答案
$$$\int \frac{x^{3}}{x^{2} - 9}\, dx = \left(\frac{x^{2}}{2} + \frac{9 \ln\left(\left|{x^{2} - 9}\right|\right)}{2}\right) + C$$$A