原函数与不定积分的性质

复习：

$$\eqalign{ & \frac{{{\text{d}}F\left( x \right)}}{{{\text{d}}x}} = f\left( x \right), \cr & {\text{d}}F\left( x \right) = f\left( x \right){\text{d}}x, \cr & \int {{\text{d}}F\left( x \right)} = \int {f\left( x \right){\text{d}}x} \cr & F\left( x \right) = \int {f\left( x \right){\text{d}}x} + C, \cr & \left[ {F\left( x \right)} \right]' = \left[ {\int {f\left( x \right){\text{d}}x} + C} \right]' \cr & \left[ {F\left( x \right)} \right]' = \left[ {\int {f\left( x \right){\text{d}}x} } \right]' = f\left( x \right) \cr} $$

性质1 设函数$f(x)$及$g(x)$的原函数存在，则

$$\int {[f\left( x \right) + g(x)]{\text{d}}x} = \int {f\left( x \right){\text{d}}x} + \int {g(x){\text{d}}x} $$

证明:等式左端是$f(x)+g(x)$的不定积分

\begin{equation} \eqalign{ & \left[ {\int {f\left( x \right){\text{d}}x} + \int {g\left( x \right){\text{d}}x} } \right]' \cr & = \left[ {\int {f\left( x \right){\text{d}}x} } \right]' + \left[ {\int {g\left( x \right){\text{d}}x} } \right]' \cr & = f\left( x \right) + g\left( x \right) \cr} \end{equation}

所以等式右端是$f(x)+g(x)$的原函数，又因为等式右端有两个不定积分符号，因此含有两个任意常数之和即一个任意常数，因此右端是$f(x)+g(x)$的原函数与一个任意常数之和，即$f(x)+g(x)$的不定积分，因此左右两端相等.

设

$$\int {[f\left( x \right) + g(x)]{\text{d}}x} = Z\left( x \right) + C$$

那么$Z' (x)=f(x)+g(x)$

设$F' (x)=f(x),G' (x)=g(x)$

则

$$\eqalign{ & \int {f\left( x \right){\text{d}}x} + \int {g(x){\text{d}}x} \cr & = F\left( x \right) + C + G\left( x \right) + C \cr & = F\left( x \right) + G\left( x \right) + C + C \cr} $$ $$\left[ {F\left( x \right) + G\left( x \right)} \right]' = F'\left( x \right) + G'\left( x \right) $$ $$ = f\left( x \right) + g(x)$$

则

$$\eqalign{ & Z'\left( x \right) - \left[ {F\left( x \right) + G\left( x \right)} \right]{'} = 0 \cr & Z\left( x \right) = F\left( x \right) + G\left( x \right) + C \cr} $$

那么，

$（1）$式左端为

$$Z(x)+C=F(x)+G(x)+C+C$$

$（1）$式右端为

$$F(x)+G(x)+C+C$$

所以左端等于右端.

 

性质2 设函数$f(x)$的原函数存在，$k$为非零常数，则

$$\int {kf(x){\text{d}}x} = k\int {f(x){\text{d}}x} $$

证明：设$F' (x)=kf(x)$，则

$$\int {kf(x){\text{d}}x} = F\left( x \right) + C$$

设$G' (x)=f(x)$，则$k\int {f(x){\text{d}}x} = kG\left( x \right) + kC$

又因为$[kG(x)]'-F' (x)=0$

所以$F(x)=kG(x)+C$

$C$是任意常数，$kC$，$C+C$也是任意常数，所以左端等于右端.

 

例 求

$$\int {\frac{{{{(x - 1)}^2}}}{x}} \text{d}x$$

解：

$$\int {\frac{{{{(x - 1)}^2}}}{x}} \text{d}x = \int {\frac{{{x^2} - 2x + 1}}{x}} \text{d}x$$ $$\eqalign{ & = \int {(x - 2 + \frac{1}{x})} \text{d}x \cr & = \int {x \text{d}x - } \int {2 \text{d}x + \int {\frac{1}{x}} \text{d}x} \cr & = \frac{1}{2}{x^2} - 2x + \ln |x| + C \cr} $$