# 分部積分法

（重定向自分部积分

## 規則

${\displaystyle {\frac {{\rm {d}}(hk)}{{\rm {d}}x}}={\frac {{\rm {d}}h}{{\rm {d}}x}}k+h{\frac {{\rm {d}}k}{{\rm {d}}x}}}$

{\displaystyle {\begin{aligned}hk&=\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{aligned}}}

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

${\displaystyle \int _{a}^{A}{\frac {{\rm {d}}h}{{\rm {d}}x}}k\ {\rm {d}}x={\big [}hk{\big ]}_{a}^{A}-\int _{a}^{A}h{\frac {{\rm {d}}k}{{\rm {d}}x}}\ {\rm {d}}x}$

${\displaystyle {\big [}hk{\big ]}_{a}^{A}=h(A)k(A)-h(a)k(a)}$

{\displaystyle {\begin{aligned}h(A)k(A)-h(a)k(a)&=\int _{a}^{A}{\frac {{\rm {d}}(hk)}{{\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}^{A}k\ {\rm {d}}h+\int _{a}^{A}h\ {\rm {d}}k\\\end{aligned}}}

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

${\displaystyle \int u\,dv=uv-\int v\,du}$

${\displaystyle \int fg\,dx=f\int g\,dx-\int \left(f'\int g\,dx\right)\,dx}$

${\displaystyle \int uv\,dw=uvw-\int uw\,dv-\int vw\,du}$

## 例题

${\displaystyle \int x\cos(x)\,dx}$

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

{\displaystyle {\begin{aligned}\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{aligned}}}

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

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

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

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

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

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

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

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

${\displaystyle \int \ln(x)\cdot 1\,dx}$

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

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

${\displaystyle \int \arctan(x)\cdot 1\,dx}$

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

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

## ILATE约法

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

${\displaystyle \int x\cos x\,dx.\,}$

${\displaystyle x\sin x-\int 1\sin x\,dx,\,}$

${\displaystyle x\sin x+\cos x+C.\,}$

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

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

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

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

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