Funtzio irrazionalen integralen zerrenda

Ondorengoa funtzio irrazionalen integralen zerrenda bat da (jatorrizkoak edo antideribatuak). Integralen zerrenda osatuago nahi baduzu, ikusi integralen zerrenda.

Bere barnean (x² +a²)-ren erroa daukaten funtzioen integralak $r = \sqrt{x^2+a^2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\int\frac{dx}{r^3} = \frac{x}{a^2r}+ K$

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

$\int\frac{x\,dx}{r^3} = -\frac{1}{r}+ K$

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

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

Bere barnean (x² -a²)-ren erroa daukaten funtzioen integralak $s = \sqrt{x^2-a^2}$

Kasu honetan $(x^2>a^2)$ izan behar da, bestela $(x^2<a^2)$ bada, begiratu hurrengo atala.

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

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

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

Kontuan hartu behar dugu $\ln\left|\frac{x+s}{a}\right| =\mathrm{sgn}(x)\cosh^{-1}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)$ dela, non $\cosh^{-1}\left|\frac{x}{a}\right|$-ren balio positiboa hartzen den.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bere barnean (a²-x²)-ren erroa daukaten funtzioen integralak $u = \sqrt{a^2-x^2}$

$\int u \;dx = \frac{1}{2}\left(xu+a^2\arcsin\frac{x}{a}\right)+ K \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int xu\;dx = -\frac{1}{3} u^3+ K \qquad\mbox{(}|x|\leq|a|\mbox{)}$

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

$\int\frac{dx}{u} = \arcsin\frac{x}{a}+ K \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{x^2\;dx}{u} = \frac{1}{2}\left(-xu+a^2\arcsin\frac{x}{a}\right)+ K \qquad\mbox{(}|x|\leq|a|\mbox{)}$

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

Bere barnean (ax²+bx+c)-ren erroa daukaten funtzioen integralak $R = \sqrt{ax^2+bx+c}$

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

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

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

$\int\frac{dx}{R} = -\frac{1}{\sqrt{-a}}\arcsin\frac{2ax+b}{\sqrt{b^2-4ac}}+ K \qquad \mbox{( }a<0\mbox{, }4ac-b^2<0\mbox{, }\left|2ax+b\right|<\sqrt{b^2-4ac}\mbox{)}$

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

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

$\int\frac{dx}{R^{2n+1}} = \frac{2}{(2n-1)(4ac-b^2)}\left(\frac{2ax+b}{R^{2n-1}}+4a(n-1)\int\frac{dx}{R^{2n-1}}\right)$

$\int\frac{x}{R}\;dx = \frac{R}{a}-\frac{b}{2a}\int\frac{dx}{R}$

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

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

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

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

Bere barnean (ax+b)-ren erroa daukaten funtzioen integralak $S = \sqrt{ax+b}$

$\int \frac{dx}{\sqrt{ax + b}}\,=\,\frac{2\sqrt{ax + b}}{a}+ K$

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

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

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

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

Bibliografia

• Peirce, Benjamin Osgood (1929), «3. atalaburua», A Short Table of Integrals (3. argitaraldia), Boston: Ginn and Co, 16.–30. or. 
• Milton Abramowitz & Irene A. Stegun, eds., Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 1972, Dover: New York. (Ikusi: capítol 3.)