MA.Add.02.积佬修炼手册

积分表

1. ${\displaystyle \int k\mathrm{d}x=kx+C(\int 0\mathrm{d}x=C)}$;

2. ${\displaystyle \int x^{\alpha }\mathrm{d}x=\frac{1}{\alpha +1}{x}^{\alpha +1}+C(\alpha \ne -1)}$;
   常用形式：

   • ${\displaystyle \int x\mathrm{d}x=\frac{1}{2}{x}^2}$
   • ${\displaystyle \int \sqrt{x}\mathrm{d}x=\frac{2}{3}{x}^{\frac{3}{2}}+C}$
   • ${\displaystyle \int \frac{\mathrm{d}x}{\sqrt{x}}=2x^{\frac{1}{2}}+C=2\sqrt{x}+C}$
   • ${\displaystyle \int \frac{\mathrm{d}x}{x^2}=-x^{-1}+C=-\frac{1}{x}+C}$

3. ${\displaystyle \int \frac{1}{x}\mathrm{d}x=\ln |x|+C}$;

   • P.S. $\ln x$ 的定义域为 $x>0$, $\frac{1}{x}$ 的定义域为 $x\ne 0$ 故需要绝对值
   • ${\displaystyle \int \frac{1}{ax+b}\mathrm{d}x=\frac{\ln |ax+b|}{a}+C}$

4. ${\displaystyle \int a^x\mathrm{d}x=\frac{a^x}{\ln a}+C(0<a\ne 1)}$;

   • P.S. $(a^x{)}^{\prime }=\ln a\cdot a^x$

5. ${\displaystyle \int {\mathrm{e}}^x\mathrm{d}x={\mathrm{e}}^x+C}$;

6. ${\displaystyle \int \sin x\mathrm{d}x=-\cos x+C}$;

7. ${\displaystyle \int \cos x\mathrm{d}x=\sin x+C}$;

8. ${\displaystyle \int {\sec }^{2}x\mathrm{d}x=\tan x+C}$;

   • ${\displaystyle \int \sec x\mathrm{d}x=\ln |\sec x+\tan x|+C}$

9. ${\displaystyle \int {\csc }^{2}x\mathrm{d}x=-\cot x+C}$;

   • ${\displaystyle \int \csc x\mathrm{d}x=-\ln |\csc x+\cot x|+C=\ln (\tan \frac{x}{2})}$

10. ${\displaystyle \int \frac{\mathrm{d}x}{a^2+x^2}=\frac{1}{a}\arctan \frac{x}{a}+C(a\ne 0)}$;

    P.S. ${(\frac{1}{a}\arctan \frac{x}{a})}^{\prime }=\frac{1}{a}\cdot \frac{1}{a}\frac{1}{1+{\left(\frac{x}{a}\right)}^2}=\frac{1}{a^2+x^2}$

    • ${\displaystyle \int \frac{\mathrm{d}x}{1+x^2}=\arctan x+C}$

11. ${\displaystyle \int \frac{\mathrm{d}x}{x^2-a^2}=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C(a\ne 0)}$;

    • ${\displaystyle \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left|\frac{x+a}{x-a}\right|+C}$

12. ${\displaystyle \int \frac{\mathrm{d}x}{\sqrt{a^2-x^2}}=\arcsin \frac{x}{a}+C(a>0)}$;

13. ${\displaystyle \int \frac{\mathrm{d}x}{\sqrt{x^2\pm a^2}}=\ln |x+\sqrt{x^2\pm a^2}|+C}$.

补充不定积分

• $\int \tan (x)\mathrm{d}x=\ln (\cos (x))$
• $\int \sqrt{x^2\pm a^2}dx=\frac{x}{2}\sqrt{x^2\pm a^2}\pm \frac{a^2}{2}\ln |x+\sqrt{x^2\pm a^2}|+C$
• $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\arcsin \frac{x}{a}+C$
• $\mathbf{\text{Reduction Formula: }}I(m,n)=\int {\cos }^{m}x{\sin }^{m}xdx=\dots$(递推)
• $\int \arctan xdx=-\frac{1}{2}\ln (x^2+1)+x\arctan (x)+C$

部分积分推导

$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan \frac{x}{a}+C$ "AA"

Tips

$(\arctan x{)}^{\prime }=\frac{1}{1+x^2}$

${(\frac{1}{a}\arctan \frac{x}{a})}^{\prime }=\frac{1}{a}\cdot \frac{1}{a}\frac{1}{1+{\left(\frac{x}{a}\right)}^2}=\frac{1}{a^2+x^2}$

Solution

$$\begin{array}{rl}& \text{Assuming }a>0\\ & =\frac{1}{a^2}\int \frac{1}{1+{\left(\frac{x}{a}\right)}^2}dx\\ & =\frac{1}{a}\int {\displaystyle \frac{d\left(\frac{x}{a}\right)}{1+{\left(\frac{x}{a}\right)}^2}}\\ & =\frac{1}{a}\arctan \frac{x}{a}+C\end{array}$$

$\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C$ 1/2a ∓

Tips

$$\text{凑2a}⟹\frac{1}{x-a}-\frac{1}{x+a}⟹{\displaystyle \frac{x-a}{x+a}}$$

Solution

$$\begin{array}{rl}& \text{Assuming }a\ne 0\\ & =\int {\displaystyle \frac{1}{(x+a)(x-a)}}dx\\ & =\frac{1}{2a}(\int \frac{1}{x-a}dx-\int \frac{1}{x+a}dx)\\ & =\frac{1}{2a}(\ln |x-a|-\ln |x+a|)+C\\ & =\frac{1}{2a}\ln \left|{\displaystyle \frac{x-a}{x+a}}\right|+C\end{array}$$

$\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln \left|\frac{x+a}{x-a}\right|+C$

Solution

类似地

$$\begin{array}{rl}& =-\int \frac{1}{x^2-a^2}dx\\ & =-\frac{1}{2a}\ln \left|{\displaystyle \frac{x-a}{x+a}}\right|+C\\ & =\frac{1}{2a}\ln \left|{\displaystyle \frac{x+a}{x-a}}\right|+C^{\prime }\end{array}$$

$\int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+C$

Tips

$(\arcsin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}}$

$(\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}$

Solution
类似 $\int \frac{1}{x^2+a^2}dx$

$$\begin{array}{r}\\ & \text{Assuming }a>0\\ & \int \frac{1}{\sqrt{a^2-x^2}}\\ & =\frac{1}{a}\int \frac{1}{\sqrt{1-{\left(\frac{x}{a}\right)}^2}}dx\\ & =\int \frac{1}{\sqrt{1-{\left(\frac{x}{a}\right)}^2}}d\left(\frac{x}{a}\right)\\ & =\arcsin \left(\frac{x}{a}\right)+C\end{array}$$

$\int \frac{1}{\sqrt{x^2\pm a^2}}dx=\int \sec tdt=\ln |\frac{x}{a}+\frac{\sqrt{x^2\pm a^2}}{a}|+C$ ▲ (b + c) / a

$\int \frac{1}{\sqrt{x^2-a^2}}dx$

Tips

#三角代换

对边：$\sqrt{x^2-a^2}$
邻边：$a$
斜边：$x$

Solution

$$\begin{array}{rl}& \int \frac{1}{\sqrt{x^2-a^2}}dx\\ & \stackrel{x=a\sec t}{=}\int \frac{1}{a\tan t}d(a\sec t)& 0<t<\frac{\pi }{2}\\ & =\int \frac{1}{\tan t}td\sec t\\ & =\int \frac{1}{\tan t}\sec t\tan tdt\\ & =\int \sec tdt\\ & =\ln |\sec t+\tan t|+C\\ & =\ln |\frac{x}{a}+{\displaystyle \frac{\sqrt{x^2-a^2}}{a}}|+C& ★\text{ 还原}\end{array}$$

$\int \frac{1}{\sqrt{x^2+a^2}}dx$

Tips

#三角代换

对边：$x$
邻边：$a$
斜边：$\sqrt{x^2+a^2}$

Solution

$$\begin{array}{rl}& \int \frac{1}{\sqrt{x^2+a^2}}dx\\ & \stackrel{x=a\tan t}{=}\int \frac{1}{a\sec t}d(a\tan t)\\ & =\int \frac{1}{\sec t}{\sec }^{2}tdx\\ & =\int \sec tdt\\ & \text{同上}\\ & =\ln |\frac{x}{a}+{\displaystyle \frac{\sqrt{x^2+a^2}}{a}}|+C& ★\text{ 还原}\end{array}$$

综上

$$\int \frac{1}{\sqrt{x^2\pm a^2}}dx=\ln |\frac{x}{a}+{\displaystyle \frac{\sqrt{x^2\pm a^2}}{a}}|+C$$

三角形两个边比a相加

$\int \tan xdx$

Solution

$$\begin{array}{rl}& =\int {\displaystyle \frac{\sin x}{\cos x}}dx\\ & =-\int \frac{1}{\cos x}d\cos x\\ & =-\ln |\cos x|+C\end{array}$$

$\int \sec xdx=\ln |\sec x+\tan x|+C$ 10'(ten sec)

Solution.I
最快，不自然

$$\begin{array}{rl}& =\int {\displaystyle \frac{\sec x\cdot (\sec x+\tan x)}{\sec x+\tan x}}dx\\ \because & (\sec x+\tan x{)}^{\prime }={\displaystyle \frac{\sin x}{{\cos }^{2}x}}+\frac{1}{{\cos }^{2}x}=\sec x\cdot (\sec x+\tan x)\\ & =\int {\displaystyle \frac{d(\sec x+\tan x)}{\sec x+\tan x}}\\ & =\ln |\sec x+\tan x|+C\end{array}$$

Solution.II
较快， #凑微分法

$$\begin{array}{rl}& =\int {\displaystyle \frac{\cos x}{{\cos }^{2}x}}dx\\ & =-\int {\displaystyle \frac{d\sin x}{1-{\sin }^{2}x}}dx\\ & =-\frac{1}{2}\ln \left(\frac{1+\sin x}{1-\sin x}\right)+C\end{array}$$

$\int \csc xdx=-\ln |\csc x+\cot x|+C$

Solution.I

$$\begin{array}{rl}& =\int -{\displaystyle \frac{\csc x\cdot -(\csc x+\cot x)}{\csc x+\cot x}}dx\\ \because & {(\csc x+\cot x)}^{\prime }=-\csc x\cdot (\csc x+\cot x)\\ & =-\int {\displaystyle \frac{d(\csc x+\cot x)}{\csc x+\cot x}}dx\\ & =-\ln |\csc x+\cot x|+C\\ & =-\ln |\tan \frac{x}{2}|\end{array}$$

Solution.II

同上，${\displaystyle =-\frac{1}{2}\ln \left({\displaystyle \frac{1+\cos x}{1-\cos x}}\right)}$

$I(m,n)=\int {\cos }^{m}x{\sin }^{n}xdx$

Solution
#分部积分法 + 递推构造

$$\begin{array}{rl}I(m,n)& =-\frac{1}{m+1}\int {\sin }^{n-1}xd({\cos }^{m+1}x)\\ & =-\frac{1}{m+1}({\sin }^{n-1}x{\cos }^{m+1}x-\int {\cos }^{m+1}xd{\sin }^{n-1}x)\dots \text{分部积分}\\ & =-\frac{1}{m+1}{\sin }^{n-1}x{\cos }^{m+1}x+{\displaystyle \frac{n-1}{m+1}}\int {\sin }^{n-2}x{\cos }^{m+2}xdx\\ & =-\frac{1}{m+1}{\sin }^{n-1}x{\cos }^{m+1}x+{\displaystyle \frac{n-1}{m+1}}{\sin }^{n-2}x{\cos }^{m}x(1-{\sin }^{2}x)dx\\ & =-\frac{1}{m+1}{\sin }^{n-1}x{\cos }^{m+1}x+{\displaystyle \frac{n-1}{m+1}}{\sin }^{n-2}x{\cos }^{m}xdx\\ & +{\displaystyle \frac{1-n}{m+1}}\int {\sin }^{n}x{\cos }^{m}xdx\\ \therefore \frac{m+n}{m+1}I(m,n)& =-\frac{1}{m+1}{\sin }^{n-1}x{\cos }^{m+1}x+{\displaystyle \frac{n-1}{m+1}}I(m,n-2)\\ ⟹I(m,n)& =-\frac{1}{m+n}{\sin }^{n-1}x{\cos }^{m+1}x+{\displaystyle \frac{n-1}{m+n}}I(m,n-2)(m\ge 0,n\ge 2)\\ \text{同理: }I(m,n)& =-\frac{1}{m+n}{\cos }^{m-1}x{\sin }^{n+1}x+{\displaystyle \frac{m-1}{m+n}}I(m-2,n)(m\ge 2,n\ge 0)\end{array}$$

Reduction Formulae

The reduction formulae for the sine function and the cosine function to an unspecific (integer) degree are:

As a particular case of their products and quotients:

Memorization

$\int {\sec }^{n}xdx$

Quiz 2 部分过程

将右边 $(n-2)\int {\sec }^{n}xdx$ 移项到左边，我们得到

也就是

再由

和

就可以求出积分。例如

典题整理

不定积分

$\int {\displaystyle \frac{\arctan x}{x^2(x^2+1)}}dx$

来源: 作业题

Analysis

#分部积分法

Hint

1. 裂项 ${\displaystyle \frac{1}{x^2(x^2+1)}=\frac{1}{x^2}-\frac{1}{x^2+1}}$
2. ${\displaystyle {(-\frac{1}{x}-\arctan x)}^{\prime }=\frac{1}{x^2}-\frac{1}{x^2+1}}$
3. ${\displaystyle \frac{1}{x^3+x}=\frac{(x^2+1)-x\cdot x}{x(x^2+1)}=\frac{1}{x}-\frac{x}{x^2+1}}$
4. ${\displaystyle \int \frac{x}{x^2+1}dx⟹\frac{1}{2}\int \frac{1}{x^2+1}d(x^2+1)}$

Solution

$\int {\displaystyle \frac{x+\sin x}{\cos x+1}}dx$

来源: 作业题

Analysis

Hint

1. 分子加法：拆分
2. #凑微分法 ${(\cos x+1)}^{\prime }=\sin x$
3. 分母 $\cos x+1$ => 二倍角/半角公式消成单一项

Solution

$I_{15}=\int {\mathrm{e}}^x(1-\frac{{\mathrm{e}}^{-x}}{\sqrt{x}})\mathrm{d}x$.

来源: 微积分每日一题3-187：求不定积分基础26题第15题

Analysis

本题考查最基本的积分表运算

Solution

$$\begin{array}{rl}{I}_{15}& =\int {\mathrm{e}}^x(1-\frac{{\mathrm{e}}^{-x}}{\sqrt{x}})\mathrm{d}x=\int ({\mathrm{e}}^x-\frac{{\mathrm{e}}^x\cdot {\mathrm{e}}^{-x}}{\sqrt{x}})\mathrm{d}x\\ & =\int {\mathrm{e}}^x\mathrm{d}x-\int \frac{1}{\sqrt{x}}\mathrm{d}x={\mathrm{e}}^x-2\int \frac{\mathrm{d}x}{2\sqrt{x}}\\ & ={\mathrm{e}}^x-2\int \mathrm{d}(\sqrt{x})={\mathrm{e}}^x-2\sqrt{x}+C.\end{array}$$

Wrong Answer

$$\int e^{x}d{\displaystyle \frac{e^{-x}}{\sqrt{x}}}\ne \int 1d\frac{1}{\sqrt{x}}$$

不能在分式上挪 $e^x$ !

#凑微分法

$\int \frac{\mathrm{d}x}{1+x^4}$

来源: 较难积分题.pdf

Analysis

运用 #添项 和分式中经典的 #凑微分法

Solution

$$\begin{array}{rl}\int \frac{\mathrm{d}x}{1+x^4}& =\frac{1}{2}\int \frac{(1+x^2)-(x^2-1)}{1+x^4}\mathrm{d}x\dots \text{经典添项}\\ & =\frac{1}{2}\int \frac{1+x^2}{1+x^4}\mathrm{d}x-\frac{1}{2}\int \frac{x^2-1}{1+x^4}\mathrm{d}x\\ & =\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\mathrm{d}x-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\mathrm{d}x\dots \text{经典凑微分}\\ & =\frac{1}{2}\int \frac{\mathrm{d}(x-\frac{1}{x})}{x^2+\frac{1}{x^2}}-\frac{1}{2}\int \frac{\mathrm{d}(x+\frac{1}{x})}{x^2+\frac{1}{x^2}}\\ & =\frac{1}{2}\int \frac{\mathrm{d}(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}-\frac{1}{2}\int \frac{\mathrm{d}(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-2}\\ & =\frac{1}{2\sqrt{2}}\arctan \frac{x-\frac{1}{x}}{\sqrt{2}}-\frac{1}{4\sqrt{2}}\ln \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+C\\ & =\frac{\sqrt{2}}{4}\arctan \frac{x^2-1}{\sqrt{2}x}-\frac{\sqrt{2}}{8}\ln \left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C.\end{array}$$

求不定积分: $I=\int \frac{\mathrm{d}x}{\sqrt[2023]{1+x^{2023}}(1+x^{2023})}$.

微积分每日一题3-109：不定积分凑微分法练习

Analysis

利用 #凑微分法 即可求解.

Solution

$$\begin{array}{rl}I& =\int \frac{\mathrm{d}x}{\sqrt[2023]{1+x^{2023}}(1+x^{2023})}=\int \frac{\mathrm{d}x}{{(1+x^{2023})}^{\frac{1}{2023}}\cdot (1+x^{2023})}\\ & =\int \frac{\mathrm{d}x}{{(1+x^{2023})}^{\frac{1}{2023}+1}}=\int \frac{\mathrm{d}x}{{\left[x^{2023}(\frac{1}{x^{2023}}+1)\right]}^{\frac{2024}{2023}}}\\ & =\int \frac{\mathrm{d}x}{{(1+x^{-2023})}^{\frac{2024}{2023}}\cdot x^{2024}}=\int \frac{x^{-2024}}{{(1+x^{-2023})}^{\frac{2024}{2023}}\mathrm{d}x}\\ & =\int (-\frac{1}{2023})\frac{1}{{(1+x^{-2023})}^{\frac{2024}{2023}}\mathrm{d}\left(x^{-2023}\right)}\\ & =-\frac{1}{2023}\int {(1+x^{-2023})}^{-\frac{2024}{2023}}\mathrm{d}(1+x^{-2023})\\ & =\frac{u=1+x^{-2023}}{-1}-\frac{1}{2023}\int u^{-\frac{2024}{2023}}\mathrm{d}u\\ & =-\frac{1}{2023}\cdot \frac{1}{-\frac{2024}{2023}+1}\cdot u^{-\frac{2024}{2023}+1}+C\\ & =-\frac{1}{2023\cdot (1-\frac{2024}{2023})}\cdot u^{-\frac{1}{2023}}+C\\ & =-\frac{1}{2023-2024}\cdot {(1+x^{-2023})}^{-\frac{1}{2023}}+C\\ & =-(-1)\cdot {(1+x^{-2023})}^{-\frac{1}{2023}}+C\\ & ={(1+x^{-2023})}^{-\frac{1}{2023}}+C.\end{array}$$

#换元积分法

${\int }_1^{+\mathrm{\infty }}\frac{1}{x\sqrt{x-1}}dx$

来源: Chap 5.4 3.(1)

Solution

$$\begin{array}{rl}& \stackrel{t=\sqrt{x-1}}{=}{\int }_0^{+\mathrm{\infty }}{\displaystyle \frac{1}{(t^2+1)t}}d(t^2+1)\\ & ={\int }_0^{+\mathrm{\infty }}\frac{1}{t^2+1}dt\\ & =2\arctan \sqrt{x-1}\end{array}$$

$\int \frac{\sqrt{x-1}\arctan \sqrt{x-1}}{x}dx$

来源: 作业题

Solution

$$\begin{array}{rl}& \int \frac{\sqrt{x-1}\arctan \sqrt{x-1}}{x}dx\\ & \text{ 设 }u=\sqrt{x-1}\text{, 则 }x=u^2+1,dx=2udu\\ & \text{ 原式 }=\int \frac{u\arctan u}{u^2+1}\cdot 2udu\\ & =2\int \frac{u^2\arctan u}{u^2+1}du\\ & =2\int \arctan udu-2\int \frac{\arctan u}{u^2+1}du\\ & =2u\arctan u-2\int \frac{u}{u^2+1}du-2\int \arctan ud(\arctan u)\\ & =2u\arctan u-\int \frac{1}{u^2+1}d\left(u^2\right)-(\arctan u{)}^2\\ & =2u\arctan u-\ln (u^2+1)-(\arctan u{)}^2+c\\ & =2\sqrt{x-1}\arctan \sqrt{x-1}-(\arctan \sqrt{x-1}{)}^2-\ln x+c\end{array}$$

🔴${\int }_0^{\frac{\pi }{4}}\sqrt{\tan x}\mathrm{d}x$

Chap 5.1 P190 22.(9)

Solution #换元积分法

$$\begin{array}{rl}& \stackrel{t=\tan x}{=}{\int }_0^1\sqrt{t}d\arctan t\\ & \stackrel{m=\sqrt{t}}{=}{\int }_0^{1}md\arctan m^2\\ & ={\int }_0^1\frac{2m^2}{1+m^4}dm\\ & ={\int }_0^1\frac{m^2-1}{1+m^4}dm+{\int }_0^1\frac{m^2+1}{1+m^{\phi }}dm\\ & ={\int }_0^1\frac{1-\frac{1}{m^2}}{m^2+\frac{1}{m^2}}dm+{\int }_0^1\frac{1+\frac{1}{m^2}}{m^2+\frac{1}{m^2}}dm\\ & ={\int }_0^1\frac{d(m+\frac{1}{m})}{m^2+\frac{1}{m^2}}+{\int }_0^1\frac{d(m-\frac{1}{m})}{m^2+\frac{1}{m^2}}\\ & ={\int }_0^1\frac{d(m+\frac{1}{m})}{{(m+\frac{1}{m})}^2-2}+{\int }_0^1\frac{d(m-\frac{1}{m})}{{(m-\frac{1}{m})}^2+2}\\ & ={\frac{1}{2\sqrt{2}}\ln \left|\frac{m+\frac{1}{m}-\sqrt{2}}{m+\frac{1}{m}+\sqrt{2}}\right||}_0^1+{\frac{1}{\sqrt{2}}\arctan \frac{m-\frac{1}{m}}{\sqrt{2}}|}_0^1\\ \because & \underset{m\to 0^{+}}{lim}(1-\frac{2\sqrt{2}}{m+\frac{1}{m}+\sqrt{2}})=1-\underset{m\to 0^{+}}{lim}\frac{2\sqrt{2}m}{m^2+\sqrt{2}m+1}=1\\ & \underset{m\to 0^{+}}{lim}(m-\frac{1}{m})\frac{0}{=}\underset{m\to 0^{+}}{lim}m-\underset{m\to 0^{+}}{lim}\frac{1}{m}=-\mathrm{\infty }\\ \therefore & =\frac{1}{2\sqrt{2}}(\ln \frac{2-\sqrt{2}}{2+\sqrt{2}}-0)+\frac{1}{\sqrt{2}}(\arctan 0-\arctan -\mathrm{\infty })\\ & =\frac{1}{2\sqrt{2}}\ln (3-2\sqrt{2})+\frac{1}{\sqrt{2}}(0+\frac{\pi }{2})\\ & =\frac{\sqrt{2}}{2}\ln (\sqrt{2}-1)+\frac{\sqrt{2}\pi }{4}\\ & \approx 0.4875\end{array}$$

反向乘积求导公式与反向商的求导公式

来自：SUDO Edu

反向乘积的导数

我们知道，两个函数的乘积的导数为

所以

例1 求积分 $\int (\frac{1}{\ln x}+\ln (\ln x))dx$.

解: 分开来算，这两个积分都算不出来。但仔细观察， $(\ln (\ln x){)}^{\prime }=\frac{1}{x\ln x}$ ，与第一项差了一个因子 $\frac{1}{x}$ ，所以将 $\ln (\ln x)$ 乘以 $x$ ，再求导，正好了被积分函数，所以

反向商的求导公式

商的求导公式为

所以

如果一个被积分函数的分母为一个函数的平方，我们可以通过商的求导公式凑出 $f(x)$ ，然后利用反向商的求导公式求出积分。

例2 求积分 $\int \frac{{\sin }^{2}x}{(x\cos x-\sin x{)}^2}dx$.

解: 因为分母是一个函数的平方，看起来还有点复杂，我们来“凑”出一个商的求导公式。因为分母为 $(x\cos x-\sin x{)}^2$ ，所以令 $g(x)=x\cos x-\sin x，g^{\prime }(x)=\cos x-x\sin x-\cos x=-x\sin x$ ，

因为右边只有 ${\sin }^{2}x$ ，一个直观的猜想是 $-f^{\prime }(x)\sin x={\sin }^{2}x$ ，也就是 $f(x)=\cos x$ ，而另外两项为 0 。将 $f(x)=\cos x$ 代入上式，

所以，原积分为

练习题

最后，给出几个习题供有兴趣的同学们练习。
求积分
(1) $\int (x{\sec }^{2}x+\tan x)dx$
(2) $\int x^x(\ln x+1)dx$
(3) $\int e^{\sin x}(x^2\cos x+2x)dx$
(4) $\int \frac{\ln x}{x^2(1-\ln x{)}^2}dx$
(5) $\int \frac{\sin x-x\cos x-\cos }{{\sin }^{2}x}$

定积分

${\int }_{-1}^1{e}^{|x|}\cdot \arctan e^{x}dx$

Solution

$$\begin{array}{rl}& (13){\int }_{-1}^1{e}^{|x|}\cdot \arctan e^{x}dx;\\ =& {\int }_{-1}^0{e}^x\arctan x{e}^{x}dx+{\int }_0^1{e}^x\arctan e^{x}dx\dots \text{分离去绝对值}\\ =& {\int }_{-1}^0{e}^{-x}(\frac{\pi }{2}-\arctan e^{-x}dx)+{\int }_0^1{e}^x\arctan e^{x}dx\\ =& \frac{\pi }{2}{\int }_{-1}^0{e}^{-x}dx-{\int }_{-1}^0{e}^{-x}\arctan e^{-x}dx+{\int }_0^1{e}^x\arctan e^{x}dx\\ =& \frac{\pi }{2}\cdot (e-1)+{\int }_{-1}^0\arctan e^{-x}d\left(e^{-x}\right)+{\int }_0^1\arctan e^{x}d\left(e^x\right).\\ =& \frac{\pi }{2}(e-1)\end{array}$$

${\int }_0^{\frac{\pi }{2}}\sqrt{\tan x}\mathrm{d}x$

Solution

$$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\sqrt{\tan x}\mathrm{d}x& =2({\int }_0^1+{\int }_1^{+\mathrm{\infty }})\frac{x^2}{1+x^4}\mathrm{d}x\\ & =2{\int }_0^1\frac{x^2}{1+x^4}\mathrm{d}x+2{\int }_0^1\frac{1}{1+x^4}\mathrm{d}x\\ & =2{\int }_0^1\frac{\mathrm{d}(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}\\ & =2{\int }_{-\mathrm{\infty }}^0\frac{\mathrm{d}x}{x^2+2}\\ & =\frac{\pi }{\sqrt{2}}.\end{array}$$

${\int }_0^{1}x\arcsin xdx$

Analysis

#分部积分法 + #三角代换

Solution

$$\begin{array}{rl}& u=\arcsin x{v}^{\prime }=x\\ & u^{\prime }=\frac{1}{\sqrt{1-x^2}}v=\frac{x^2}{2}\\ & \int x\arcsin xdx=\frac{x^2}{2}\arcsin x-\frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}dx\text{. }\\ \\ & \mathbf{\text{第二类换元 }}\\ & \text{ 故 }x=\sin t.\Rightarrow dx=\cos tdt\\ & (t=\arcsin x),x\in [0,1]\Rightarrow t\in [0,\frac{\pi }{2}]\\ \\ & \int \frac{x^2}{\sqrt{1-x^2}}dx\\ & \stackrel{x=\sin t}{=}\int {\displaystyle \frac{{\sin }^{2}t}{\cos t}}d\sin t\\ & =\int {\sin }^{2}tdt\\ & =\int (\frac{1}{2}-\frac{1}{2}\cos 2t)dt\\ & =\frac{1}{2}t-\frac{\sin 2t}{4}+C\\ \therefore & \int x\arcsin xdx\\ & ={\frac{x^2}{2}\arcsin x|}_0^1-{\frac{1}{2}[\frac{1}{2}t-\frac{\sin 2t}{4}]|}_0^{\pi /2}\\ & =\frac{\pi }{4}-\frac{1}{2}(\frac{\pi }{4}-0)\text{. }\\ & =\frac{\pi }{8}\\ & \end{array}$$