高数强化NO13|不定积分的计算|原函数|不定积分|公式|换元积分|分部积分

基本概念

原函数

$$\begin{array}{rl}& \text{设函数}F(x)\text{在区间}I\text{上可导，对区间}I\text{上的每一点都有}{F}^{\prime }(x)=f(x)\text{，则称函数}\\ & F(x)\text{是}f(x)\text{在区间}I\text{上的一个原函数.}\\ & \\ & \boxed{【注】}①\text{提到原函数一般要指明区间，不加特别说明的情况下一般就默认为}f(x)\text{的定}\\ & \text{义域.}\\ & \\ & ②\text{如果}f(x)\text{在区间}I\text{上存在原函数，那么它的原函数一定不唯一.}\end{array}$$

不定积分

$$\begin{array}{rl}& \text{函数}f(x)\text{在区间}I\text{上的所有原函数组成的集合，称为}f(x)\text{在区间}I\text{上的不定积分，}\\ & \text{记作}\int f(x)dx.\\ & \\ & \boxed{【注】}\\ & 1.\text{原函数就是导数的反义词：}f(x)\text{是}F(x)\text{的导数}⟺F(x)\text{是}f(x)\text{的原函数.}\\ & \text{对于原函数的概念，要注意两点：一是提到原函数需要指明区间，二是原函数只要存在}\\ & \text{就不是唯一的.}\\ & \\ & 2.f(x)\text{在区间}I\text{上的不定积分是}f(x)\text{在该区间上所有原函数组成的集合.}\\ & \text{要注意概念所揭示的不定积分的本质：它是一个集合.}\\ & \\ & 3.\text{不定积分和原函数的关系：假设}F(x)\text{是}f(x)\text{在区间}I\text{上的一个原函数，}\int f(x)dx\text{是}\\ & f(x)\text{在区间}I\text{上的不定积分，则}F(x)\in \int f(x)dx,\int f(x)dx=F(x)+C\\ & \\ & 4.\text{任意常数}C\text{的运算法则：}C+C=C-C=C,CC=C,kC=\{\begin{array}{l}C,k\ne 0\\ 0,k=0\end{array}\end{array}$$

基本性质

$$\begin{array}{rl}& \text{设}f(x),g(x)\text{均存在原函数，则}\\ & (1)\int [f(x)\pm g(x)]dx=\int f(x)dx\pm \int g(x)dx；\\ & \\ & (2)\int kf(x)dx=k\int f(x)dx(k\in R,k\ne 0)；\\ & \\ & (3){\left(\int f(x)dx\right)}^{\prime }=f(x)\text{或}d\int f(x)dx=f(x)dx；\\ & \\ & (4)\int F^{\prime }(x)dx=F(x)+C\text{或}\int dF(x)=F(x)+C.\\ & \\ & \boxed{【注】}①\text{当}k=0\text{时，}\int kf(x)dx=\int 0dx=C\text{，而}k\int f(x)dx=0\cdot [F(x)+C]=0\text{，两}\\ & \text{边是不相等的，所以要注意第二个公式中的}k\ne 0.②\text{不定积分满足线性性质.}\end{array}$$

基本方法

积分公式

$$\begin{array}{rl}& (1)\int x^{a}dx=\frac{1}{a+1}{x}^{a+1}+C(a\ne -1),\int \frac{1}{x}dx=\ln |x|+C；\\ & \\ & (2)\int a^{x}dx=\frac{1}{\ln a}{a}^x+C(a>0\text{且}a\ne 1),\int e^{x}dx=e^x+C；\\ & \\ & (3)\int \cos xdx=\sin x+C,\int \sin xdx=-\cos x+C；\\ & \\ & (4)\int {\sec }^{2}xdx=\tan x+C,\int {\csc }^{2}xdx=-\cot x+C；\\ & \\ & (5)\int \sec x\tan xdx=\sec x+C,\int \csc x\cot xdx=-\csc x+C；\\ & \\ & (6)\int \frac{1}{1+x^2}dx=\arctan x+C；\\ & \\ & (7)\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C；\\ & \\ & (8)\int \sec xdx=\ln |\sec x+\tan x|+C,\int \csc xdx=\ln |\csc x-\cot x|+C；\\ & \\ & (9)\int \frac{1}{a^2{x}^2+b^2}dx=\frac{1}{ab}\arctan \frac{ax}{b}+C,\int \frac{1}{\sqrt{b^2-a^2{x}^2}}dx=\frac{1}{a}\arcsin \frac{ax}{b}+C；\\ & \\ & (10)\int \frac{dx}{\sqrt{x^2\pm a^2}}=\ln \left|x+\sqrt{x^2\pm a^2}\right|+C\end{array}$$

换元积分法

1. 第一类换元法
   $$\begin{array}{rl}& \text{设}f(u)\text{有一个原函数}F(u)，u=\phi (x)\text{可导，则有}\\ & \int f[\phi (x)]{\phi }^{\prime }(x)dx\stackrel{\text{令}u=\phi (x)}{=}\int f(u)du=F(u)+C=F[\phi (x)]+C.\end{array}$$
2. 第二类换元法
   $$\begin{array}{rl}& \text{设}x=\psi (t)\text{是单调，可导的函数，并且}{\psi }^{\prime }(t)\ne 0,\text{设}f[\psi (t)]{\psi }^{\prime }(t)\text{具有原函数}G(t),\\ & \text{则}\int f(x)dx\stackrel{x=\psi (t)}{=}\int f[\psi (t)]{\psi }^{\prime }(t)dt=G(t)+C=G\left[{\psi }^{-1}(x)\right]+C.\\ & \\ & \boxed{【注】}①x=\psi (t)\text{必须存在反函数；}②\text{注意最后结果回代，用}x\text{表示.}\end{array}$$

分部积分法

$$\begin{array}{rl}& \text{分部积分公式：}\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx\text{，或简写为}\int udv=uv-\int vdu.\end{array}$$

对不定积分概念和性质的考查

$$\begin{array}{rl}& f^{\prime }(x)=2(x-1),f(x)=(x-1{)}^2+C,x\in [0,2],f(0)=0,C=-1\\ & \\ & f(x)=x^2-2x,x\in [0,2],f(7)=f(3)=f(-1)=-f(1)=1\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{不定积分是有定义域的，在求原函数时，一定要指明区间。}\end{array}$$

$$选B$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{积分号遇到微分号，或是微分号遇到积分号，直接“湮灭”（抵消）；}\\ & 2.\text{求积分要带}C,\text{求导不用带}C(\text{原因：}C\text{求导就为}0)。\end{array}$$

不定积分的计算

$$\begin{array}{rl}& \text{解: 令}x=\tan t\\ & \int \frac{\tan t+1}{(1+{\tan }^{2}t{)}^2}d\tan t=\int \frac{\tan t+1}{{\sec }^{4}t}{\sec }^{2}tdt=\int (\sin t\cos t+{\cos }^{2}t)dt\\ & =\frac{1}{2}{\sin }^{2}t+\int \frac{1+\cos 2t}{2}dt=\frac{1}{2}{\sin }^{2}t+\frac{1}{2}t+\frac{\sin 2t}{4}+C\\ & =\frac{1}{2}{\sin }^{2}t+\frac{1}{2}t+\frac{1}{2}\sin t\cos t+C\\ & =\frac{1}{2}{\left(\frac{x}{\sqrt{1+x^2}}\right)}^2+\frac{1}{2}t+\frac{1}{2}\left(\frac{x}{\sqrt{1+x^2}}\right)\left(\frac{1}{\sqrt{1+x^2}}\right)+C\\ & =\frac{x^2+x}{2(1+x^2)}+\frac{1}{2}\arctan x+C\end{array}$$
$$\begin{array}{rl}& =\frac{x^2+1+x-1}{2(1+x^2)}+\frac{1}{2}\arctan x+C\\ & =\frac{x-1}{2(1+x^2)}+\frac{1}{2}\arctan x+C+\frac{1}{2}\\ & =\frac{x-1}{2(1+x^2)}+\frac{1}{2}\arctan x+C\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{如果分母为}{\left(a^2+x^2\right)}^n\text{，也可以考虑做变量三角代换}x=a\tan t。\end{array}$$

$$\begin{array}{rl}& \text{解: (1)法1: }\int {\tan }^{3}x{\sec }^{2}x{\sec }^{2}xdx\\ & =\int {\tan }^{3}x(1+{\tan }^{2}x)d\tan x\\ & =\frac{1}{4}{\tan }^{4}x+\frac{1}{6}{\tan }^{6}x+C\\ & \\ & \text{法2: }\int {\tan }^{2}x{\sec }^{3}x(\tan x\sec x)dx\\ & =\int ({\sec }^{2}x-1){\sec }^{3}xd\sec x\\ & =\frac{1}{6}{\sec }^{6}x-\frac{1}{4}{\sec }^{4}x+C\end{array}$$
$$\begin{array}{rl}\int \sec xdx& =\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x}dx\\ & =\int \frac{d(\tan x+\sec x)}{\sec x+\tan x}\\ & =\ln |\sec x+\tan x|+C\end{array}$$
$$\begin{array}{rl}& (2)\int {\sec }^{3}xdx\\ & =\int \sec x{\sec }^{2}xdx\\ & =\int \sec xd\tan x\\ & =\sec x\tan x-\int \tan xd\sec x\\ & =\sec x\tan x-\int \sec x{\tan }^{2}xdx\\ & =\sec x\tan x-\int \sec x({\sec }^{2}x-1)dx\\ & =\sec x\tan x-\int {\sec }^{3}xdx+\int \sec xdx\\ & \\ & \int {\sec }^{3}xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\int \sec xdx\\ & =\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|+C\end{array}$$
$$\begin{array}{rl}\int {\sec }^{5}xdx& =\int {\sec }^{3}xd\tan x\\ & ={\sec }^{3}x\tan x-\int \tan x\cdot 3{\sec }^{2}x\cdot \sec x\tan xdx\\ & ={\sec }^{3}x\tan x-3\int {\sec }^{3}x({\sec }^{2}x-1)dx\\ & \\ \int {\sec }^{5}xdx& =\frac{1}{4}{\sec }^{3}x\tan x+\frac{3}{4}\int {\sec }^{3}xdx\end{array}$$
$$\begin{array}{rl}& (3)\int {\tan }^{2}x{\sec }^{3}xdx\\ & =\int ({\sec }^{2}x-1){\sec }^{3}xdx\\ & =\int ({\sec }^{5}x-{\sec }^{3}x)dx\\ & =\frac{1}{4}{\sec }^{3}x\tan x-\frac{1}{4}\int {\sec }^{3}xdx\\ & =\frac{1}{4}{\sec }^{3}x\tan x-\frac{1}{8}\sec x\tan x-\frac{1}{8}\ln |\sec x+\tan x|+C\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{对于形如}\int {\tan }^{m}x{\sec }^{n}xdx\text{的三角有理式积分，可以按照如下方法计算：}\\ & 1.\text{如果}m\text{为奇数或}n\text{为偶数，则可以按照如下方法凑微分：}\\ & \text{若}m=2k+1\text{，则}\\ & \int {\tan }^{2k+1}x{\sec }^{n}xdx=\int {\tan }^{2k}x{\sec }^{n-1}x\tan x\sec xdx=\int {\left({\sec }^{2}x-1\right)}^k{\sec }^{n-1}xd\sec x；\\ & \text{若}n=2k\text{，则}\\ & \int {\tan }^{m}x{\sec }^{2k}xdx=\int {\tan }^{m}x{\sec }^{2k-2}x{\sec }^{2}xdx=\int {\tan }^{m}x{\left({\tan }^{2}x+1\right)}^{k-1}d\tan x。\\ & \\ & 2.\text{对于}m\text{为偶数且}n\text{为奇数，则可以按如下方法计算：}\\ & \text{若}m=0\text{，则通过分部积分对}\sec x\text{降次：}\\ & \begin{array}{rl}\int {\sec }^{2k+1}xdx& =\int {\sec }^{2k-1}x{\sec }^{2}xdx\\ & =\int {\sec }^{2k-1}xd\tan x\\ & =\tan x{\sec }^{2k-1}x-\int \tan xd{\sec }^{2k-1}x\\ & =\tan x{\sec }^{2k-1}x-(2k-1)\int {\sec }^{2k-1}x{\tan }^{2}xdx\\ & =\tan x{\sec }^{2k-1}x-(2k-1)\int {\sec }^{2k+1}xdx+(2k-1)\int {\sec }^{2k-1}xdx\end{array}\\ & \\ & \text{若}m=2l\text{，则将}{\tan }^{2l}x\text{写成}{\left({\sec }^{2}x-1\right)}^l\text{，再将括号打开，将原式化为若干个}\sec x\text{的奇}\\ & \text{数次幂的积分再进行计算。}\end{array}$$

$$\begin{array}{rl}& \text{解: 令}t=\sqrt{\frac{2-x}{x}},x=\frac{2}{t^2+1}\\ & \text{则原式}=\int \ln (1+t)d\left(\frac{2}{t^2+1}-1\right)=\ln (1+t)\left(\frac{2}{t^2+1}-1\right)-\int \frac{1}{1+t}\cdot \left(\frac{2}{t^2+1}-1\right)dt\\ & =\ln (1+t)\cdot \frac{1-t^2}{t^2+1}-\int \frac{1}{1+t}\cdot \frac{1-t^2}{t^2+1}dt=\ln (1+t)\frac{1-t^2}{t^2+1}+\int \frac{t-1}{t^2+1}dt\\ & =\ln (1+t)\frac{1-t^2}{t^2+1}+\frac{1}{2}\ln (t^2+1)-\frac{1}{2}\arctan t+C\\ & =\ln \left(1+\sqrt{\frac{2-x}{x}}\right)(x-1)+\frac{1}{2}\ln \frac{2}{x}-\frac{1}{2}\arctan \sqrt{\frac{2-x}{x}}+C\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{运用分部积分法在计算}dv\text{时，往往需要结合简单的凑微分的技巧。}\end{array}$$

$$\begin{array}{rl}& \text{解: }\int e^{2x}({\tan }^{2}x+2\tan x+1)dx\\ & =\int e^{2x}({\sec }^{2}x+2\tan x)dx\\ & =\int \left(e^{2x}(\tan x{)}^{\prime }+(e^{2x}{)}^{\prime }\tan x\right)dx\\ & =\int (e^{2x}\tan x{)}^{\prime }dx=e^{2x}\tan x+C\end{array}$$

$$\begin{array}{rl}& \text{解: }\int \frac{e^x(x-1{)}^2}{(1+x^2{)}^2}dx\\ & =\int \frac{e^x(x^2+1)-e^x\cdot 2x}{(1+x^2{)}^2}dx\\ & =\int \frac{(e^x{)}^{\prime }\cdot (x^2+1)-e^x\cdot (x^2+1{)}^{\prime }}{(1+x^2{)}^2}dx\\ & =\int {\left(\frac{e^x}{x^2+1}\right)}^{\prime }dx=\frac{e^x}{x^2+1}+C\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{当被积函数比较复杂，难于直接计算时，可以考虑将被积函数分成两部分，凑}\\ & \text{成}u{v}^{\prime }+u^{\prime }v\text{，再直接运用公式}\int (u{v}^{\prime }+u^{\prime }v)dx=uv+C\text{来进行计算。}\end{array}$$