$$$x^{2} + \operatorname{asin}{\left(x \right)}$$$ 的积分
您的输入
求$$$\int \left(x^{2} + \operatorname{asin}{\left(x \right)}\right)\, dx$$$。
解答
逐项积分:
$${\color{red}{\int{\left(x^{2} + \operatorname{asin}{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{x^{2} d x} + \int{\operatorname{asin}{\left(x \right)} d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=2$$$:
$$\int{\operatorname{asin}{\left(x \right)} d x} + {\color{red}{\int{x^{2} d x}}}=\int{\operatorname{asin}{\left(x \right)} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\int{\operatorname{asin}{\left(x \right)} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$
对于积分$$$\int{\operatorname{asin}{\left(x \right)} d x}$$$,使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=\operatorname{asin}{\left(x \right)}$$$ 和 $$$\operatorname{dv}=dx$$$。
则 $$$\operatorname{du}=\left(\operatorname{asin}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{\sqrt{1 - x^{2}}}$$$ (步骤见 »),并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。
所以,
$$\frac{x^{3}}{3} + {\color{red}{\int{\operatorname{asin}{\left(x \right)} d x}}}=\frac{x^{3}}{3} + {\color{red}{\left(\operatorname{asin}{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{\sqrt{1 - x^{2}}} d x}\right)}}=\frac{x^{3}}{3} + {\color{red}{\left(x \operatorname{asin}{\left(x \right)} - \int{\frac{x}{\sqrt{1 - x^{2}}} d x}\right)}}$$
设$$$u=1 - x^{2}$$$。
则$$$du=\left(1 - x^{2}\right)^{\prime }dx = - 2 x dx$$$ (步骤见»),并有$$$x dx = - \frac{du}{2}$$$。
因此,
$$\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} - {\color{red}{\int{\frac{x}{\sqrt{1 - x^{2}}} d x}}} = \frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} - {\color{red}{\int{\left(- \frac{1}{2 \sqrt{u}}\right)d u}}}$$
对 $$$c=- \frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$:
$$\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} - {\color{red}{\int{\left(- \frac{1}{2 \sqrt{u}}\right)d u}}} = \frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} - {\color{red}{\left(- \frac{\int{\frac{1}{\sqrt{u}} d u}}{2}\right)}}$$
应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=- \frac{1}{2}$$$:
$$\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{2}=\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{2}=\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{2}=\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{2}=\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{2}$$
回忆一下 $$$u=1 - x^{2}$$$:
$$\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \sqrt{{\color{red}{u}}} = \frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \sqrt{{\color{red}{\left(1 - x^{2}\right)}}}$$
因此,
$$\int{\left(x^{2} + \operatorname{asin}{\left(x \right)}\right)d x} = \frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \sqrt{1 - x^{2}}$$
加上积分常数:
$$\int{\left(x^{2} + \operatorname{asin}{\left(x \right)}\right)d x} = \frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \sqrt{1 - x^{2}}+C$$
答案
$$$\int \left(x^{2} + \operatorname{asin}{\left(x \right)}\right)\, dx = \left(\frac{x^{3}}{3} + x \operatorname{asin}{\left(x \right)} + \sqrt{1 - x^{2}}\right) + C$$$A