分部积分法

不定积分是被积函数的所有原函数的集合，而根据牛顿-莱布尼兹公式，定积分可以看做原函数的增量，因此，如果函数$u(x)$和$v(x)$在区间$[a,b]$上可导，则

$$[u(x)v(x)]'=u' (x)v(x)+u(x) v' (x)$$

得

$$u(x) v' (x)=[u(x)v(x)]'-u' (x)v(x)$$

两端在区间$[a,b]$上求不定积分，得

$$\eqalign{ & \int_a^b {u(x)v'(x){\text{d}}x} = \left[ {\int {u(x)v'(x){\text{d}}x} } \right]_a^b \cr & = \left[ {u(x)v(x) - \int_a^b {v(x)u'(x){\text{d}}x} } \right]_a^b \cr & = \left[ {u(x)v(x)} \right]_a^b - \int_a^b {v(x)u'(x){\text{d}}x} \cr} $$

简记为

$$\int_a^b {uv'{\text{d}}x} = \left[ {uv} \right]_a^b - \int_a^b {vu'{\text{d}}x} $$

或

$$\int_a^b {u{\text{d}}v} = [uv]_a^b - \int_a^b {v{\text{d}}u} $$

这就是定积分的分部积分公式.

例 求

$$\int_0^1 {3{{\text{e}}^{\sqrt x }}{\text{d}}x} .$$

解 令$\sqrt x = t$，则$x=t^2$,${\text{d}}x = 2t{\text{d}}t$

且当$x=0$时，$t=0$；当$x=1$时，$t=1$.

则

$$\eqalign{ & \int_0^1 {3{{\text{e}}^{\sqrt x }}{\text{d}}x} = 6\int_0^1 {t{{\text{e}}^t}{\text{d}}t} = 6\int_0^1 {t{\text{d}}({{\text{e}}^t})} \cr & = 6\left( {\left[ {t{{\text{e}}^t}} \right]_0^1 - \int_0^1 {{{\text{e}}^t}{\text{d}}t} } \right) = 6({\text{e}} - \left[ {{{\text{e}}^t}} \right]_0^1) \cr & = 6[{\text{e}} - ({\text{e}} - 1)] = 6 \cr} $$