ملحق:قائمة تكاملات الدوال غير النسبية

فيما يلي قائمة لتكاملات التوابع غير المنطقة

توابع تحتوي $r = \sqrt{a^2+x^2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

توابع تحتوي $s = \sqrt{x^2-a^2}$

بفرض أن $(x^2>a^2)$, for $(x^2<a^2)$, انظر المقطع اللاحق :

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

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

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

لاحظ أن $\ln\left|\frac{x+s}{a}\right| =\mathrm{sgn}(x)\cosh^{-1}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)$, حيث القيمة الموجبة ل$\cosh^{-1}\left|\frac{x}{a}\right|$ يجب أن تكون.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

توابع تحتوي $t = \sqrt{a^2-x^2}$

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

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

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

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

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

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

توابع تحتوي $R^{1/2} = \sqrt{ax^2+bx+c}$

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

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

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

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

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

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

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

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

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

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

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

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

توابع تحتوي $R^{1/2} = \sqrt{ax+b}$

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

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

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

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