无理函数积分表

以下是部分无理函数的积分表。(书写时省略了不定积分结果中都含有的任意常数Cn)

## 包括 r = √a2 + x2的积分

$$\int r \;dx = \frac{1}{2}\left[x r +a^2\,\ln\left(\frac{x+r}{a}\right)\right]$$

$$\int r^3 \;dx = \frac{1}{4}xr^3+\frac{1}{8}3a^2xr+\frac{3}{8}a^4\ln\left(\frac{x+r}{a}\right)$$

$$\int r^5 \; dx = \frac{1}{6}xr^5+\frac{5}{24}a^2xr^3+\frac{5}{16}a^4xr+\frac{5}{16}a^6\ln\left(\frac{x+r}{a}\right)$$

$$\int x r\;dx=\frac{r^3}{3}$$

$$\int x r^3\;dx=\frac{r^5}{5}$$

$$\int x r^{2n+1}\;dx=\frac{r^{2n+3}}{2n+3}$$

$$\int x^2 r\;dx= \frac{xr^3}{4}-\frac{a^2xr}{8}-\frac{a^4}{8}\ln\left(\frac{x+r}{a}\right)$$

$$\int x^2 r^3\;dx= \frac{xr^5}{6}-\frac{a^2xr^3}{24}-\frac{a^4xr}{16}-\frac{a^6}{16}\ln\left(\frac{x+r}{a}\right)$$

$$\int x^3 r \; dx = \frac{r^5}{5} - \frac{a^2 r^3}{3}$$

$$\int x^3 r^3 \; dx = \frac{r^7}{7}-\frac{a^2r^5}{5}$$

$$\int x^3 r^{2n+1} \; dx = \frac{r^{2n+5}}{2n+5} - \frac{a^2 r^{2n+3}}{2n+3}$$

$$\int x^4 r\;dx= \frac{x^3r^3}{6}-\frac{a^2xr^3}{8}+\frac{a^4xr}{16}+\frac{a^6}{16}\ln\left(\frac{x+r}{a}\right)$$

$$\int x^4 r^3\;dx= \frac{x^3r^5}{8}-\frac{a^2xr^5}{16}+\frac{a^4xr^3}{64}+\frac{3a^6xr}{128}+\frac{3a^8}{128}\ln\left(\frac{x+r}{a}\right)$$

$$\int x^5 r \; dx = \frac{r^7}{7} - \frac{2 a^2 r^5}{5} + \frac{a^4 r^3}{3}$$

$$\int x^5 r^3 \; dx = \frac{r^9}{9} - \frac{2 a^2 r^7}{7} + \frac{a^4 r^5}{5}$$

$$\int x^5 r^{2n+1} \; dx = \frac{r^{2n+7}}{2n+7} - \frac{2a^2r^{2n+5}}{2n+5}+\frac{a^4 r^{2n+3}}{2n+3}$$

$$\int\frac{r\;dx}{x} = r-a\ln\left|\frac{a+r}{x}\right| = r - a \sinh^{-1}\frac{a}{x}$$

$$\int\frac{r^3\;dx}{x} = \frac{r^3}{3}+a^2r-a^3\ln\left|\frac{a+r}{x}\right|$$

$$\int\frac{r^5\;dx}{x} = \frac{r^5}{5}+\frac{a^2r^3}{3}+a^4r-a^5\ln\left|\frac{a+r}{x}\right|$$

$$\int\frac{r^7\;dx}{x} = \frac{r^7}{7}+\frac{a^2r^5}{5}+\frac{a^4r^3}{3}+a^6r-a^7\ln\left|\frac{a+r}{x}\right|$$

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

$$\int\frac{x\,dx}{r} = r$$

$$\int\frac{x^2\;dx}{r} = \frac{x}{2}r-\frac{a^2}{2}\,\sinh^{-1}\frac{x}{a} = \frac{x}{2}r-\frac{a^2}{2}\ln\left|x+r\right|$$

$$\int\frac{dx}{xr} = -\frac{1}{a}\,\sinh^{-1}\frac{a}{x} = -\frac{1}{a}\ln\left|\frac{a+r}{x}\right|$$

## 包括 s = √x2 - a2的积分

假设 $(x^2>a^2)$, $(x^2<a^2)$请看下一节:

$$\int xs\;dx = \frac{1}{3}s^3$$

$$\int\frac{s\;dx}{x} = s - a\cos^{-1}\left|\frac{a}{x}\right|$$

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

注意在$\ln\left|\frac{x+s}{a}\right| =\mathrm{sgn}(x)\cosh^{-1}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)$, 里$\cosh^{-1}\left|\frac{x}{a}\right|$取正值.

$$\int\frac{x\;dx}{s} = s$$

$$\int\frac{x\;dx}{s^3} = -\frac{1}{s}$$

$$\int\frac{x\;dx}{s^5} = -\frac{1}{3s^3}$$

$$\int\frac{x\;dx}{s^7} = -\frac{1}{5s^5}$$

$$\int\frac{x\;dx}{s^{2n+1}} = -\frac{1}{(2n-1)s^{2n-1}}$$

$$\int\frac{x^{2m}\;dx}{s^{2n+1}} = -\frac{1}{2n-1}\frac{x^{2m-1}}{s^{2n-1}}+\frac{2m-1}{2n-1}\int\frac{x^{2m-2}\;dx}{s^{2n-1}}$$

$$\int\frac{x^2\;dx}{s} = \frac{xs}{2}+\frac{a^2}{2}\ln\left|\frac{x+s}{a}\right|$$

$$\int\frac{x^2\;dx}{s^3} = -\frac{x}{s}+\ln\left|\frac{x+s}{a}\right|$$

$$\int\frac{x^4\;dx}{s} = \frac{x^3s}{4}+\frac{3}{8}a^2xs+\frac{3}{8}a^4\ln\left|\frac{x+s}{a}\right|$$

$$\int\frac{x^4\;dx}{s^3} = \frac{xs}{2}-\frac{a^2x}{s}+\frac{3}{2}a^2\ln\left|\frac{x+s}{a}\right|$$

$$\int\frac{x^4\;dx}{s^5} = -\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}+\ln\left|\frac{x+s}{a}\right|$$

$$\int\frac{x^{2m}\;dx}{s^{2n+1}} = (-1)^{n-m}\frac{1}{a^{2(n-m)}}\sum_{i=0}^{n-m-1}\frac{1}{2(m+i)+1}{n-m-1 \choose i}\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\qquad\mbox{(}n>m\ge0\mbox{)}$$

$$\int\frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}$$

$$\int\frac{dx}{s^5}=\frac{1}{a^4}\left[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}\right]$$

$$\int\frac{dx}{s^7} =-\frac{1}{a^6}\left[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}\right]$$

$$\int\frac{dx}{s^9} =\frac{1}{a^8}\left[\frac{x}{s}-\frac{3}{3}\frac{x^3}{s^3}+\frac{3}{5}\frac{x^5}{s^5}-\frac{1}{7}\frac{x^7}{s^7}\right]$$

$$\int\frac{x^2\;dx}{s^5}=-\frac{1}{a^2}\frac{x^3}{3s^3}$$

$$\int\frac{x^2\;dx}{s^7} = \frac{1}{a^4}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}\right]$$

$$\int\frac{x^2\;dx}{s^9} = -\frac{1}{a^6}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}\right]$$

## 包括 t = √a2 - x2的积分

$$\int t \;dx = \frac{1}{2}\left(xt+a^2\sin^{-1}\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}$$

$$\int xt\;dx = -\frac{1}{3} t^3 \qquad\mbox{(}|x|\leq|a|\mbox{)}$$

$$\int\frac{t\;dx}{x} = t-a\ln\left|\frac{a+t}{x}\right| \qquad\mbox{(}|x|\leq|a|\mbox{)}$$

$$\int\frac{dx}{t} = \sin^{-1}\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}$$

$$\int\frac{x^2\;dx}{t} = -\frac{x}{2}t+\frac{a^2}{2}\sin^{-1}\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}$$

$$\int t\;dx = \frac{1}{2}\left(xt-\sgn x\,\cosh^{-1}\left|\frac{x}{a}\right|\right) \qquad\mbox{(}|x|\ge|a|\mbox{)}$$

$$\int{x^n\sqrt{1-x^2}\text{d}x}=\frac{\left( x^{n+1}-x^{n-1} \right) \sqrt{1-x^2}}{n+1}+\frac{n-1}{n+1}\int{x^{n-2}\sqrt{1-x^2}\text{d}x}$$

## 包括√R = √ax2 + bx + c的积分

$$\int\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a R}+2ax+b\right| \qquad\qquad \mbox{(}a>0\mbox{)}$$

$$\int\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}}\,\sinh^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\qquad \mbox{(}a>0\mbox{, }4ac-b^2>0\mbox{)}$$

$$\int\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}}\ln|2ax+b| \quad \mbox{(}a>0\mbox{, }4ac-b^2=0\mbox{)}$$

$$\int\frac{dx}{\sqrt{ax^2+bx+c}} = -\frac{1}{\sqrt{-a}}\arcsin\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad\qquad \mbox{(}a<0\mbox{, }4ac-b^2<0\mbox{)}$$

$$\int\frac{dx}{\sqrt{(ax^2+bx+c)^{3}}} = \frac{4ax+2b}{(4ac-b^2)\sqrt{R}}$$

$$\int\frac{dx}{\sqrt{(ax^2+bx+c)^{5}}} = \frac{4ax+2b}{3(4ac-b^2)\sqrt{R}}\left(\frac{1}{R}+\frac{8a}{4ac-b^2}\right)$$

$$\int\frac{dx}{\sqrt{(ax^2+bx+c)^{2n+1}}} = \frac{4ax+2b}{(2n-1)(4ac-b^2)R^{(2n-1)/2}}+\frac{8a(n-1)}{(2n-1)(4ac-b^2)} \int\frac{dx}{R^{(2n-1)/2}}$$

$$\int\frac{x\;dx}{\sqrt{ax^2+bx+c}} = \frac{\sqrt{R}}{a}-\frac{b}{2a}\int\frac{dx}{\sqrt{R}}$$

$$\int\frac{x\;dx}{\sqrt{(ax^2+bx+c)^3}} = -\frac{2bx+4c}{(4ac-b^2)\sqrt{R}}$$

$$\int\frac{x\;dx}{\sqrt{(ax^2+bx+c)^{2n+1}}} = -\frac{1}{(2n-1)aR^{(2n-1)/2}}-\frac{b}{2a}\int\frac{dx}{R^{(2n+1)/2}}$$

$$\int\frac{dx}{x\sqrt{ax^2+bx+c}}=-\frac{1}{\sqrt{c}}\ln\left(\frac{2\sqrt{c R}+bx+2c}{x}\right)$$

$$\int\frac{dx}{x\sqrt{ax^2+bx+c}}=-\frac{1}{\sqrt{c}}\sinh^{-1}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right)$$

## 包括 √R = √ax + b的积分

$$\int \frac{dx}{x\sqrt{ax + b}}\,=\,\frac{-2}{\sqrt{b}}\tanh^{-1}{\sqrt{\frac{ax + b}{b}}}$$

$$\int\frac{\sqrt{ax + b}}{x}\,dx\;=\;2\left(\sqrt{ax + b} - \sqrt{b}\tanh^{-1}{\sqrt{\frac{ax + b}{b}}}\right)$$

$$\int\frac{x^n}{\sqrt{ax + b}}\,dx\;=\;\frac{2}{a(2n+1)} \left(x^{n}\sqrt{ax + b} - bn\int\frac{x^{n-1}}{\sqrt{ax + b}}\right)$$

$$\int x^n \sqrt{ax + b}\,dx \; = \; \frac{2}{2n +1}\left(x^{n+1} \sqrt{ax + b} + bx^{n} \sqrt{ax + b} - nb\int x^{n-1}\sqrt{ax + b}\,dx \right)$$