Listo de integraloj de malracionalaj funkcioj

Jen estas listo de nedifinitaj integraloj (malderivaĵoj) de malracionalaj funkcioj.

Integraloj kun $r=\sqrt{a^2+x^2}$

$\int rdx=\frac{1}{2}(xr+a^2\ln \left(\frac{x+r}{a}\right))$

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

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

$\int xrdx=\frac{r^3}{3}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\int \frac{xdx}{r}=r$

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

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

Integraloj kun $s=\sqrt{x^2-a^2}$

Ĉi tie estas supozite ke $(x^2>a^2)$, por $(x^2<a^2)$ vidu la sekvan sekcion:

$\int xsdx=\frac{1}{3}{s}^3$

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

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

Notu, ke $\ln \left|\frac{x+s}{a}\right|=\mathrm{s}\mathrm{g}\mathrm{n}(x){\cosh }^{-1}\left|\frac{x}{a}\right|=\frac{1}{2}\ln \left(\frac{x+s}{x-s}\right)$, kie la pozitiva valoro de ${\cosh }^{-1}\left|\frac{x}{a}\right|$ estas prenata.

$\int \frac{xdx}{s}=s$

$\int \frac{xdx}{s^3}=-\frac{1}{s}$

$\int \frac{xdx}{s^5}=-\frac{1}{3s^3}$

$\int \frac{xdx}{s^7}=-\frac{1}{5s^5}$

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

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

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

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

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

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

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

$\int \frac{x^{2m}dx}{s^{2n+1}}=(-1{)}^{n-m}\frac{1}{a^{2(n-m)}}\sum _{i=0}^{n-m-1}\frac{1}{2(m+i)+1}(\genfrac{}{}{0ex}{}{n-m-1}{i})\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\text{(}n>m\ge 0\text{)}$

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

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

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

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

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

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

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

Integraloj kun $t=\sqrt{a^2-x^2}$

Ĉi tie estas supozite ke $(a^2>x^2)$, por $(a^2<x^2)$ vidu la antaŭan sekcion:

$\int tdx=\frac{1}{2}(xt+a^2{\sin }^{-1}\frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int xtdx=-\frac{1}{3}{t}^3\text{(}|x|\le |a|\text{)}$

$\int \frac{tdx}{x}=t-a\ln \left|\frac{a+t}{x}\right|\text{(}|x|\le |a|\text{)}$

$\int \frac{dx}{t}={\sin }^{-1}\frac{x}{a}\text{(}|x|\le |a|\text{)}$

$\int \frac{x^{2}dx}{t}=-\frac{x}{2}t+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\text{(}|x|\le |a|\text{)}$

$\int tdx=\frac{1}{2}(xt-\mathrm{sgn}x{\cosh }^{-1}\left|\frac{x}{a}\right|)\text{(por }|x|\ge |a|\text{)}$

Integraloj kun $R^{1/2}=\sqrt{a{x}^2+bx+c}$

$\int \frac{dx}{\sqrt{a{x}^2+bx+c}}=\frac{1}{\sqrt{a}}\ln |2\sqrt{aR}+2ax+b|\text{(por }a>0\text{)}$

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

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

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

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

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

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

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

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

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

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

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

Integraloj kun $R^{1/2}=\sqrt{ax+b}$

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

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

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

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