不定积分¶

不定积分的概念和运算法则¶

不定积分,微分的逆运算¶

定义 6.1.1¶

在某个区间上,函数\(F(x)\)和\(f(x)\)成立关系:\(F'(x)=f(x)\),或者等价的\(dF(x)=f(x)dx\),则称\(F(x)\)是\(f(x)\)的一个原函数

定义 6.1.2¶

一个函数\(f(x)\)的原函数全体称为这个函数的不定积分,用\(\int f(x)dx\)表示

不定积分的线性性质¶

定理 6.1.1¶

若函数\(f(x)\)和\(g(x)\)都在区间\(I\)上有原函数,则\(\forall k,l\in R\),函数\(kf+lg\)的原函数也存在,并且:\(\int(kf(x)+lg(x))dx=k\int f(x)dx+l\int g(x)dx\)

换元积分法和分部积分法¶

换元积分法¶

第一类换元积分法¶

\(F(x)\)是\(f(x)\)的一个原函数,则\(\int f(\varphi(x))\varphi'(x)dx=F(\varphi(x))+C\)

第二类换元积分法¶

\(F(t)\)是\(f(\varphi(t))\varphi'(t)\)的一个原函数,且\(x = \varphi(t)\),则\(\int f(x)dx = \int f(\varphi(t))\varphi'(t)dt = F(\varphi^{-1}(x))+C\)

分部积分法¶

\(\int u(x)v'(x)dx = u(x)v(x) - \int u'(x)v(x)dx\)

基本积分表¶

\(\begin{aligned} & \int x^\alpha \mathrm{d} x = \begin{cases} \frac{1}{\alpha+1} x^{\alpha+1}+C, & \alpha \neq -1, \\ \ln |x|+C, & \alpha=-1 ; \end{cases} & \int \ln x \mathrm{~d} x = x(\ln x-1)+C ; \\ & \int a^x \mathrm{~d} x = \frac{a^x}{\ln a}+C, \text{ 特别地 } \int \mathrm{e}^x \mathrm{~d} x = \mathrm{e}^x+C ; & \int \sin x \mathrm{~d} x = -\cos x+C ; \\ & \int \cos x \mathrm{~d} x = \sin x+C \text{;} & \int \tan x \mathrm{~d} x = -\ln |\cos x|+C ;\\ & \int \cot x \mathrm{~d} x = \ln |\sin x|+C \text{;} & \int \sec x \mathrm{~d} x = \ln |\sec x+\tan x|+C ; \\ & \int \csc x \mathrm{~d} x = \ln |\csc x-\cot x|+C ; & \int \operatorname{sh} x \mathrm{~d} x = \operatorname{ch} x+C ; \\ & \int \operatorname{ch} x \mathrm{~d} x = \operatorname{sh} x+C ; & \int \frac{\mathrm{d} x}{\sqrt{a^2-x^2}} = \arcsin \frac{x}{a}+C ; \\ & \int \frac{\mathrm{d} x}{\sqrt{x^2 \pm a^2}} = \ln \left|x+\sqrt{x^2 \pm a^2}\right|+C ; & \int \frac{\mathrm{d} x}{x^2-a^2} = \frac{1}{2 a} \ln \left|\frac{x-a}{x+a}\right|+C ; \\ & \int \frac{\mathrm{d} x}{x^2+a^2} = \frac{1}{a} \arctan \frac{x}{a}+C ; & \int \sqrt{a^2-x^2} \mathrm{~d} x = \frac{1}{2} x \sqrt{a^2-x^2}+\frac{a^2}{2} \arcsin \frac{x}{a}+C ; \\ & \int \sqrt{x^2 \pm a^2} \mathrm{~d} x = \frac{1}{2}\left(x \sqrt{x^2 \pm a^2} \pm a^2 \ln \left|x+\sqrt{x^2 \pm a^2}\right|\right)+C . \end{aligned}\)

一些证明:
\[\int \sec(x)dx = \int \frac{1}{\cos(x)}dx = \int \frac{1}{\cos^2(x)}d\sin(x) \overset{t=\sin(x)}{=} \int \frac{1}{2}\ln(|\frac{t+1}{t-1}|) = \ln(|\frac{1+\sin(x)}{\cos(x)}|) = \ln(|\sec(x)+\tan(x)|) +C\]
\[\int \frac{dx}{\sqrt{a^2-x^2}} = \int \frac{dx}{\sqrt{a^2(1-\frac{x^2}{a^2})}} = \int \frac{dx}{a\sqrt{1-(\frac{x}{a})^2}} \overset{t=\frac{x}{a}}{=} \int \frac{dt}{\sqrt{1-t^2}} = \arcsin(t)+C = \arcsin(\frac{x}{a}) + C\]
\[\int\frac{dx}{\sqrt{x^2\pm a^2}} = \int\frac{dx}{\sqrt{a^2(1\pm\frac{x^2}{a^2})}} = \int\frac{dx}{a\sqrt{1\pm(\frac{x}{a})^2}} \overset{t=\frac{x}{a}}{=} \int\frac{dt}{\sqrt{1\pm t^2}} = \ln(|t+\sqrt{1+t^2}|) = \ln(|\frac{x}{a}+\sqrt{1+(\frac{x}{a})^2}|) + C = \ln(|x+\sqrt{x^2\pm a^2}|) + C\]
\[\int\sqrt{a^2-x^2}dx\overset{x = a\cos(\theta)}{=} -\int |a^2\sin^2(\theta)d\theta = -a^2\int \frac{1-2\cos(2\theta)}{2}d\theta = -\frac{1}{2}a^2\theta+\frac{1}{2}\sin(\theta)\cos(\theta)+C = -\frac{1}{2}a^2\cos(\frac{x}{a})+\frac{1}{2}\frac{x}{a}\sin(\frac{x}{a}) + C= \frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}\arcsin(\frac{x}{a})+C\]
\[\begin{aligned} & \sqrt{x^2-a^2}=a \tan t, \Rightarrow d x=a \sec t \tan t d t , \\ & \int \sqrt{x^2-a^2} d x=a^2 \int \sec t \tan ^2 t d t=a^2 \int \sec t\left(\sec ^2 t-1\right) d t \\ & =a^2 \int \sec ^3 t d t-a^2 \int \sec t d t \\ & =a^2 \int \sec ^3 t d t-a^2 \ln |\sec t+\tan t|+C_1 \\ & \int \sec ^3 t=\int \frac{d(\sin t)}{\left(1-\sin ^2 t\right)^2}=\frac{1}{4} \int\left(\frac{1}{1+\sin t}+\frac{1}{1-\sin t}\right)^2 d(\sin t) \\ & =\frac{1}{4} \int \frac{d(1+\sin t)}{(1+\sin t)^2}-\frac{1}{4} \int \frac{d(1-\sin t)}{(1-\sin t)^2}+\frac{1}{2} \int \frac{d(\sin t)}{1-\sin ^2 t} \\ & =\frac{1}{4}\left(\frac{1}{1-\sin t}-\frac{1}{1+\sin t}\right)+\frac{1}{4} \ln \left(\frac{1+\sin t}{1-\sin t}\right)+C_2 \\ & =\frac{1}{2} \tan t \sec t+\frac{1}{2} \ln (\sec t+\tan t)+C_2 \\ & \int \sqrt{x^2-a^2} d x=\frac{a^2}{2} \tan t \sec t+\frac{a^2}{2} \ln |\sec t+\tan t|-a^2 \ln |\sec t+\tan t|+C \\ & =\frac{a^2}{2} \tan t \sec t-\frac{a^2}{2} \ln |\sec t+\tan t|+C \\ & =\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \ln \left|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}\right|+C \end{aligned}\]
\(\int\sqrt{x^2+a^2}dx\)同理

有理函数的不定积分及其应用¶

定理 6.3.1¶

设有理函数\(\frac{p(x)}{q(x)}\)是真分式,多项式\(q(x)\)有\(k\)重实数根\(\alpha,i.e.\exists q_1(x),s.t.q(x)=(x-\alpha)^kq_1(x)\),则\(\exists \lambda\in\mathbb{R},p_1(x)\in\mathbb{Q}[x],\deg p_1<\deg p,s.t.\)
- \[\frac{p(x)}{q(x)}=\frac{\lambda}{(x-\alpha)^k}+\frac{p_1(x)}{(x-\alpha)^{k-1}q_1(x)}\]

定理 6.3.1¶

设有理函数\(\frac{p(x)}{q(x)}\)是真分式,多项式\(q(x)\)有\(l\)重共轭复根\(\beta\pm i\gamma,i.e.\exists q_1(x),s.t.q(x)=(x+2\xi x+\eta^2)^lq_1(x),\xi=-\beta,\eta = \beta^2+\gamma^2\),则\(\exists \mu,\nu\in\mathbb{R},p_1(x)\in\mathbb{Q}[x],\deg p_1<\deg p,s.t.\)
- \[\frac{p(x)}{q(x)}=\frac{\mu x+\nu}{(x^2+2\xi x+\eta^2)^l}+\frac{p_1(x)}{(x^2+2\xi x+\eta^2)^{l-1}q_1(x)}\]

[!NOTE] - 重复利用上述定理,任意有理分式可以化为一些真分式的和,这些和形如 - \(\(\frac{\lambda_1}{x-\alpha}+\frac{\lambda_2}{(x-\alpha)^2}+\cdots+\frac{\lambda_k}{(x-\alpha)^k}\)\) - 若\(\alpha\)是\(k\)重实数根 - \(\(\frac{\mu_1 x+\nu_1}{x^2+2\xi x+\eta^2}+\frac{\mu_2 x+\nu_2}{(x^2+2\xi x+\eta^2)^2}+\cdots+\frac{\mu_l x+\nu_l}{(x^2+2\xi x+\eta^2)^l}\)\) - 若\(\beta\pm i\gamma\)是\(l\)重共轭复根