笔记下载：分部积分法
1，\((uv)’=u’v+uv’\) \(\qquad\ \ \)\(\left(\int F'(x)dx=F(x)+c\right)\)
\(\quad\)\(\int(uv)’dx=\int u’v+\int uv’\)\(\quad\Longrightarrow\quad\)\(uv=\int u’v+\int uv’\quad\Longrightarrow\int uv’dx=uv-\int u’vdx\)
2，幂函数与三角函数、指数函数的积
例1a，求\(\displaystyle\int xe^x dx\)
解答：令\(u=x\)，\(e^x=v’\)，\(v=e^x\)
\(\displaystyle\int xe^xdx=xe^x-\int 1\cdot e^xdx=xe^x-e^x+C\)
例1b，求\(\displaystyle\int x^2\sin xdx\)
解答：令\(u=x^2\)，\(v’=\sin x\)，\(v=\cos x\)，
\(\displaystyle\int x^2\sin xdx=-x^2\cos x-\int 2x(-\cos x)dx\qquad\left(u=x, v’=\cos x, v=\sin x\right)\)
\(\displaystyle\qquad\ \ =-x^2\cos x +2x\sin x-2\int 1\cdot\sin xdx\)
\(\displaystyle\qquad\ \ =-x^2\cos x+2x\sin x+2\cos x+C\)
3，幂函数与对数函数、反三角函数的乘积
幂函数： \(v’\)
例2a，求\(\displaystyle\int x\ln xdx\)
解答：\(\displaystyle=\frac{1}{2} x^2\ln x- \int \frac{1}{2} x^2 \cdot\frac{1}{x} dx\)
\(\displaystyle\quad\ \ =\frac{1}{2} x^2\ln x – \int \frac{1}{2} xdx\)
\(\displaystyle\quad\ \ =\frac{1}{2} x^2\ln x – \frac{1}{4} x^2+C\)
例3，求\(\displaystyle\int \arctan xdx\)
解答：\(\displaystyle=\int 1\cdot \arctan xdx\)
\(\displaystyle\quad\ \ =x \arctan x-\int \frac{x}{1+x^2} dx\)
\(\displaystyle\quad\ \ =x \arctan x-\frac{1}{2}\ln \left(1+x^2\right)+C\)
4，三角函数与指数函数的积 （回复积分）
例3，\(\displaystyle\int e^x \sin x dx=e^x\sin x-\int e^x\cdot\cos x dx\)
解答：=\(\displaystyle e^x \sin x -\left(e^x\cdot\cos x-\int e^x\cdot(-\sin x)\right)dx\)
\(\displaystyle\quad\ \ =e^x \sin x-\left(e^x\cos x-\int e^x\cdot(-\sin x)\right)dx\)
\(\displaystyle\quad\ \ =e^x\sin x-e^x\cos x -\int e^x \sin xdx\)
\(\displaystyle\Longrightarrow2\int e^x\sin x dx=e^x\sin x-e^x\cos x+C\)
\(\displaystyle\Longrightarrow \int e^x\sin x dx=\frac{1}{2} e^x \sin x- \frac{1}{2} e^x\cos x+C\)