# 分部積分法

$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega$

## 规则

$\frac{{\rm{d}}h\!k}{{\rm{d}}x}=\frac{{\rm{d}}h}{{\rm{d}}x}\!k\!+\!h\!\frac{{\rm{d}}k}{{\rm{d}}x}$

\begin{align} h\!k &= \int\frac{{\rm{d}}h}{{\rm{d}}x}\!k\!+\!h\!\frac{{\rm{d}}k}{{\rm{d}}x}\ {\rm{d}}\!x\\ &= \int h\ {\rm{d}}\!k+\int k\ {\rm{d}}\!h\\ \end{align}

$\int\frac{{\rm{d}}h}{{\rm{d}}x}\!k\ {\rm{d}}\!x=h\!k-\int h\!\frac{{\rm{d}}k}{{\rm{d}}x}\ {\rm{d}}\!x$

$\int^A_a\frac{{\rm{d}}h}{{\rm{d}}x}\!k\ {\rm{d}}\!x=\big[h\!k\big]^A_a-\int^A_ah\!\frac{{\rm{d}}k}{{\rm{d}}x}\ {\rm{d}}\!x$

$\big[h\!k\big]^A_a=h(A)k(A)-h(a)k(a)$

\begin{align} h(A)k(A)-h(a)k(a) &= \int^A_a\frac{{\rm{d}}h\!k}{{\rm{d}}x}\ dx\\ & = \int^A_a\frac{{\rm{d}}h}{{\rm{d}}x}\!k\!+\!h\!\frac{{\rm{d}}k}{{\rm{d}}x}\ dx\\ & = \int^A_ak\ {\rm{d}}\!h+\int^A_ah\ {\rm{d}}\!k\\ \end{align}

$\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,$

$\int u\,dv=uv-\int v\,du$

$\int f g\,dx = f \int g\,dx - \int \left ( f' \int g\,dx \right )\,dx$

$\int u v\,dw = u v w - \int u w\,dv - \int v w\,du$

## 例题

$\int x\cos (x) \,dx$

u = x,故du = dx,
dv = cos(x) dx,故v = sin(x).

\begin{align} \int x\cos (x) \,dx & = \int u \,dv \\ & = uv - \int v \,du \\ & = x\sin (x) - \int \sin (x) \,dx \\ & = x\sin (x) + \cos (x) + C \end{align}

$\int x^{3} \sin (x) \,dx \quad \mbox{and} \quad \int x^{2} e^{x} \,dx$

$\int e^{x} \cos (x) \,dx$

u = cos(x);故du = −sin(x) dx
dv = ex dx;故v = ex

$\int e^{x} \cos (x) \,dx = e^{x} \cos (x) + \int e^{x} \sin (x) \,dx$

u = sin(x); du = cos(x) dx
v = ex; dv = ex dx

 $\int e^{x} \sin (x) \,dx$ $= e^{x} \sin (x) - \int e^{x} \cos (x) \,dx$

$\int e^{x} \cos (x) \,dx = e^{x} \cos (x) + e^x \sin (x) - \int e^{x} \cos (x) \,dx$

$2 \int e^{x} \cos (x) \,dx = e^{x} ( \sin (x) + \cos (x) ) + C$
$\int e^{x} \cos (x) \,dx = {e^{x} ( \sin (x) + \cos (x) ) \over 2} + C$

$\int \ln (x) \cdot 1 \,dx$

u = ln(x); du = 1/x dx
v = x; dv = 1·dx

 $\int \ln (x) \,dx$ $= x \ln (x) - \int \frac{x}{x} \,dx$ $= x \ln (x) - \int 1 \,dx$
$\int \ln (x) \,dx = x \ln (x) - {x} + {C}$
$\int \ln (x) \,dx = x ( \ln (x) - 1 ) + C$

$\int \arctan (x) \cdot 1 \,dx$

u = arctan(x); du = 1/(1+x2) dx
v = x; dv = 1·dx

 $\int \arctan (x) \,dx$ $= x \arctan (x) - \int \frac{x}{1 + x^2} \,dx$ $= x \arctan (x) - {1 \over 2} \ln \left( 1 + x^2 \right) + C$

## ILATE约法

I: 反三角函数：arctan x, arcsec x, etc.
L: 对数函数：ln x, $\log_2(x)$, etc.
A: 代数函数$x^2$, $3x^{50}$, etc.
T: 三角函数：sin x, tan x, etc.
E: 指数函数$e^x$, $13^x$, etc.

u确定后，另一个函数自然是dv. ILATE这个口诀代表优先选择的顺序。.其中的道理是求列在后面的函数的积分比列在前面的更容易。

$\int x\cos x \,dx.\,$

$x\sin x - \int 1\sin x \,dx,\,$

$x\sin x + \cos x+C. \,$

$\frac{x^2}2\cos x + \int \frac{x^2}2\sin x\,dx\,\,$

ILATE约法尽管很有用，也还是会有例外。所以有时可以用"LIATE"顺序替换。另外，在个别情况要将指数项拆开。例如，求积分

$\int x^3e^{x^2}\,dx,$

$u=x^2,\quad dv=xe^{x^2}\,dx$

$\int x^3e^{x^2}\,dx=\frac12e^{x^2}(x^2-1)+C$