$$$\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1$$$ 的積分
您的輸入
求$$$\int \left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x}\right)}}$$
配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$\int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{x}}$$
由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:
$$x + {\color{red}{\int{\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} d x}}} = x + {\color{red}{\int{\left(x^{2} - 2 + \frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}}\right)d x}}}$$
逐項積分：
$$x + {\color{red}{\int{\left(x^{2} - 2 + \frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}}\right)d x}}} = x + {\color{red}{\left(- \int{2 d x} + \int{x^{2} d x} + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x}\right)}}$$
配合 $$$c=2$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$x + \int{x^{2} d x} + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} - {\color{red}{\int{2 d x}}} = x + \int{x^{2} d x} + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} - {\color{red}{\left(2 x\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：
$$- x + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\int{x^{2} d x}}}=- x + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- x + \int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
進行部分分式分解（步驟可見 »）:
$$\frac{x^{3}}{3} - x + {\color{red}{\int{\frac{3 x^{2} + 2}{\left(x^{2} + 1\right)^{2}} d x}}} = \frac{x^{3}}{3} - x + {\color{red}{\int{\left(\frac{3}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}}}$$
逐項積分：
$$\frac{x^{3}}{3} - x + {\color{red}{\int{\left(\frac{3}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}}} = \frac{x^{3}}{3} - x + {\color{red}{\left(- \int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x} + \int{\frac{3}{x^{2} + 1} d x}\right)}}$$
為了計算積分 $$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}$$$，對積分 $$$\int{\frac{1}{x^{2} + 1} d x}$$$ 使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
令 $$$\operatorname{u}=\frac{1}{x^{2} + 1}$$$ 與 $$$\operatorname{dv}=dx$$$。
則 $$$\operatorname{du}=\left(\frac{1}{x^{2} + 1}\right)^{\prime }dx=- \frac{2 x}{\left(x^{2} + 1\right)^{2}} dx$$$（步驟見 »），且 $$$\operatorname{v}=\int{1 d x}=x$$$（步驟見 »）。
因此，
$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}=\frac{1}{x^{2} + 1} \cdot x-\int{x \cdot \left(- \frac{2 x}{\left(x^{2} + 1\right)^{2}}\right) d x}=\frac{x}{x^{2} + 1} - \int{\left(- \frac{2 x^{2}}{\left(x^{2} + 1\right)^{2}}\right)d x}$$
將常數提出：
$$\frac{x}{x^{2} + 1} - \int{\left(- \frac{2 x^{2}}{\left(x^{2} + 1\right)^{2}}\right)d x}=\frac{x}{x^{2} + 1} + 2 \int{\frac{x^{2}}{\left(x^{2} + 1\right)^{2}} d x}$$
將被積分函數的分子改寫為 $$$x^{2}=x^{2}{\color{red}{+1}}{\color{red}{-1}}$$$，並拆分：
$$\frac{x}{x^{2} + 1} + 2 \int{\frac{x^{2}}{\left(x^{2} + 1\right)^{2}} d x}=\frac{x}{x^{2} + 1} + 2 \int{\left(- \frac{1}{\left(x^{2} + 1\right)^{2}} + \frac{x^{2} + 1}{\left(x^{2} + 1\right)^{2}}\right)d x}=\frac{x}{x^{2} + 1} + 2 \int{\left(\frac{1}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}$$
將積分拆分：
$$\frac{x}{x^{2} + 1} + 2 \int{\left(\frac{1}{x^{2} + 1} - \frac{1}{\left(x^{2} + 1\right)^{2}}\right)d x}=\frac{x}{x^{2} + 1} - 2 \int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x} + 2 \int{\frac{1}{x^{2} + 1} d x}$$
因此，我們得到如下關於該積分的簡單線性方程：
$$\int{\frac{1}{x^{2} + 1} d x}=\frac{x}{x^{2} + 1} + 2 \int{\frac{1}{x^{2} + 1} d x} - 2 {\color{red}{\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}}}$$
解得
$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}=\frac{x}{2 \left(x^{2} + 1\right)} + \frac{\int{\frac{1}{x^{2} + 1} d x}}{2}$$
因此，
$$\frac{x^{3}}{3} - x + \int{\frac{3}{x^{2} + 1} d x} - {\color{red}{\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}}} = \frac{x^{3}}{3} - x + \int{\frac{3}{x^{2} + 1} d x} - {\color{red}{\left(\frac{x}{2 \left(x^{2} + 1\right)} + \frac{\int{\frac{1}{x^{2} + 1} d x}}{2}\right)}}$$
$$$\frac{1}{x^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$：
$$\frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} + \int{\frac{3}{x^{2} + 1} d x} - \frac{{\color{red}{\int{\frac{1}{x^{2} + 1} d x}}}}{2} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} + \int{\frac{3}{x^{2} + 1} d x} - \frac{{\color{red}{\operatorname{atan}{\left(x \right)}}}}{2}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=3$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{2} + 1}$$$：
$$\frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + {\color{red}{\int{\frac{3}{x^{2} + 1} d x}}} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + {\color{red}{\left(3 \int{\frac{1}{x^{2} + 1} d x}\right)}}$$
$$$\frac{1}{x^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$：
$$\frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + 3 {\color{red}{\int{\frac{1}{x^{2} + 1} d x}}} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} - \frac{\operatorname{atan}{\left(x \right)}}{2} + 3 {\color{red}{\operatorname{atan}{\left(x \right)}}}$$
因此，
$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} + \frac{5 \operatorname{atan}{\left(x \right)}}{2}$$
化簡：
$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)}$$
加上積分常數：
$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)}+C$$
答案
$$$\int \left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)\, dx = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)} + C$$$A