分部积分法
\[\begin{aligned} &\int udv=uv-\int vdu. \end{aligned}\]
- 目标: 求出原函数.
- 策略: 交换\(u,v\)化简积分.
- 重点: 何时用?怎么用?
1 何时用?
被积函数求导后简单. 如 \[\begin{aligned} &\ln x, \arctan x, \arcsin x. \end{aligned}\]
经过一次或两次分部积分后, 又得到要求的不定积分, 出现循环. 通过解方程得到不定积分. 如 \[\begin{aligned} &\int e^x \sin {x}dx. \end{aligned}\]
2 怎么用?
- 关键是找到合适的\(v\).
求 \(\displaystyle\int x \cdot e^x dx.\)
\[ \int x \cdot e^x dx = x \cdot e^x - \int e^x \cdot dx. \]
求 \(\displaystyle\int \ln x dx.\)
\[\begin{aligned} &\int \ln x dx \\ =& x \cdot \ln x - \int x \cdot \frac{1}{x} dx \\ =& x \ln x - \int 1 dx. \end{aligned}\]
求 \(\displaystyle\int \arcsin x \, dx.\)
\[\begin{aligned} &\int \arcsin x \, dx \\ =& x \cdot \arcsin x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} dx. \end{aligned}\]
求 \(\displaystyle\int x^2 e^x \, dx.\)
\[\begin{aligned} &\int x^2 e^x dx =\int x^2 de^x \\ =& x^2 e^x - \int e^x \cdot 2x dx. \end{aligned}\]
求 \(\displaystyle\int e^x \sin x \, dx.\)
\[\begin{aligned} &\int e^x \sin x dx =\int \sin x de^x\\ =&e^x \sin x - \int e^x \cos x dx \\ =&e^x \sin x - \int \cos x de^x\\ =&e^x \sin x -e^x\cos x-\int e^x\sin x dx. \end{aligned}\]
- 移项求解