Primitivele funcţiilor iraţionale

De la Wikipedia, enciclopedia liberă

Acest articol face parte din seria de articole
Primitive ale diferitelor funcţii

Tabel de integrale

Raţionale

Logaritmice

Exponenţiale

Iraţionale

Trigonometrice

Hiperbolice

Invers trigonometrice

Hiperbolice reciproce

Următorul articol este o listă de integrale (primitive) de funcţii iraţionale. Pentru o listă cu mai multe integrale, vezi tabel de integrale şi lista integralelor.

[modifică] Integrale cu $r = \sqrt{x^2+a^2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

[modifică] Integrale cu $s = \sqrt{x^2-a^2}$

Se presupune $(x 2 > a 2)$ , pentru $(x 2 < a 2)$ , vezi următoarea secţiune:

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

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

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

$\ln\left|\frac{x+s}{a}\right| =\mathrm{sgn}(x)\cosh^{-1}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)$ , unde se consideră valoarea pozitivă a lui $\cosh^{-1}\left|\frac{x}{a}\right|$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

[modifică] Integrale cu $t = \sqrt{a^2-x^2}$

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

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

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

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

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

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

[modifică] Integrale cu $R^{1/2} = \sqrt{ax^2+bx+c}$

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

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

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

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

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

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

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

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

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

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

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

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

[modifică] Integrale cu $R^{1/2} = \sqrt{ax+b}$

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

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

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

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

Adus de la "http://ro.wikipedia.org../../../p/r/i/Primitivele_func%C5%A3iilor_ira%C5%A3ionale.html"

Categorii: Analiză matematică | Calcul diferenţial şi integral | Integrale

We provide Linux to the World

Primitivele funcţiilor iraţionale

De la Wikipedia, enciclopedia liberă

Cuprins

[modifică] Integrale cu $r = \sqrt{x^2+a^2}$

[modifică] Integrale cu $s = \sqrt{x^2-a^2}$

[modifică] Integrale cu $t = \sqrt{a^2-x^2}$

[modifică] Integrale cu $R^{1/2} = \sqrt{ax^2+bx+c}$

[modifică] Integrale cu $R^{1/2} = \sqrt{ax+b}$

Views

Navigare

Caută

În alte limbi