Zerrenda:Funtzio arrazionalen integralak

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

$\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+K\text{( }n\ne -1\text{)}$

$\int \frac{c}{ax+b}dx=\frac{c}{a}\ln |ax+b|+K$

$\int x(ax+b{)}^{n}dx=\frac{a(n+1)x-b}{a^2(n+1)(n+2)}(ax+b{)}^{n+1}+K\text{( }n\in ̸\{-1,-2\}\text{)}$

$\int \frac{x}{ax+b}dx=\frac{x}{a}-\frac{b}{a^2}\ln |ax+b|+K$

$\int \frac{x}{(ax+b{)}^2}dx=\frac{b}{a^2(ax+b)}+\frac{1}{a^2}\ln |ax+b|+K$

$\int \frac{x}{(ax+b{)}^n}dx=\frac{a(1-n)x-b}{a^2(n-1)(n-2)(ax+b{)}^{n-1}}+K\text{( }n\in ̸\{1,2\}\text{)}$

$\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+K$

$\int \frac{x^2}{ax+b}dx=\frac{b^2\ln (|ax+b|)}{a^3}+\frac{a{x}^2-2bx}{2a^2}+K$

$\int \frac{x^2}{(ax+b{)}^2}dx=\frac{1}{a^3}(ax-2b\ln |ax+b|-\frac{b^2}{ax+b})+K$

$\int \frac{x^2}{(ax+b{)}^3}dx=\frac{1}{a^3}(\ln |ax+b|+\frac{2b}{ax+b}-\frac{b^2}{2(ax+b{)}^2})+K$

$\int \frac{x^2}{(ax+b{)}^n}dx=\frac{1}{a^3}(-\frac{(ax+b{)}^{3-n}}{(n-3)}+\frac{2b(ax+b{)}^{2-n}}{(n-2)}-\frac{b^2(ax+b{)}^{1-n}}{(n-1)})+K\text{( }n\in ̸\{1,2,3\}\text{)}$

$\int \frac{1}{x(ax+b)}dx=-\frac{1}{b}\ln \left|\frac{ax+b}{x}\right|+K$

$\int \frac{1}{x^2(ax+b)}dx=-\frac{1}{bx}+\frac{a}{b^2}\ln \left|\frac{ax+b}{x}\right|+K$

$\int \frac{1}{x^2(ax+b{)}^2}dx=-a(\frac{1}{b^2(ax+b)}+\frac{1}{a{b}^{2}x}-\frac{2}{b^3}\ln \left|\frac{ax+b}{x}\right|)+K$

$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan \frac{x}{a}+K$

$\int \frac{1}{x^2-a^2}dx=\{\begin{array}{cc}-\frac{1}{a}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\frac{x}{a}=\frac{1}{2a}\ln \frac{a-x}{a+x}+K& \text{(for }|x|<|a|\text{)}\\ -\frac{1}{a}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{h}\frac{x}{a}=\frac{1}{2a}\ln \frac{x-a}{x+a}+K& \text{( }|x|>|a|\text{)}\end{array}$

$a\ne 0$ bada:

$\int \frac{1}{a{x}^2+bx+c}dx=\{\begin{array}{cc}\frac{2}{\sqrt{4ac-b^2}}\arctan \frac{2ax+b}{\sqrt{4ac-b^2}}+K& \text{( }4ac-b^2>0\text{)}\\ -\frac{2}{\sqrt{b^2-4ac}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\frac{2ax+b}{\sqrt{b^2-4ac}}+K=\frac{1}{\sqrt{b^2-4ac}}\ln \left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|+K& \text{( }4ac-b^2<0\text{)}\\ -\frac{2}{2ax+b}+K& \text{( }4ac-b^2=0\text{)}\end{array}$

$\int \frac{x}{a{x}^2+bx+c}dx=\frac{1}{2a}\ln |a{x}^2+bx+c|-\frac{b}{2a}\int \frac{dx}{a{x}^2+bx+c}$

$\int \frac{mx+n}{a{x}^2+bx+c}dx=\{\begin{array}{cc}\frac{m}{2a}\ln |a{x}^2+bx+c|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan \frac{2ax+b}{\sqrt{4ac-b^2}}+K& \text{( }4ac-b^2>0\text{)}\\ \frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\frac{2ax+b}{\sqrt{b^2-4ac}}+K& \text{( }4ac-b^2<0\text{)}\\ \frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a(2ax+b)}+K& \text{( }4ac-b^2=0\text{)}\end{array}$

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

$\int \frac{x}{(a{x}^2+bx+c{)}^n}dx=-\frac{bx+2c}{(n-1)(4ac-b^2)(a{x}^2+bx+c{)}^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int \frac{1}{(a{x}^2+bx+c{)}^{n-1}}dx$

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

$\int \frac{dx}{x^{2^n}+1}=\sum _{k=1}^{2^{n-1}}\{\frac{1}{2^{n-1}}[\sin \left(\frac{(2k-1)\pi }{2^n}\right)\arctan [(x-\cos \left(\frac{(2k-1)\pi }{2^n}\right))\csc \left(\frac{(2k-1)\pi }{2^n}\right)]]-\frac{1}{2^n}[\cos \left(\frac{(2k-1)\pi }{2^n}\right)\ln |x^2-2x\cos \left(\frac{(2k-1)\pi }{2^n}\right)+1|]\}$

Edozein funtzio arrazional integra dezakegu goiko berdintzak erabiliz eta zatiki arrazionalen integrazioaren artikuluan eskaintzen diren teknikak aplikatuz, funtzio arrazionalak ondorengo formako batugaietan banatzearen bidez:

$\frac{ex+f}{{(a{x}^2+bx+c)}^n}$.