Seznam integralov racionalnih funkcij

Seznam integralov racionalnih funkcij Naslednji seznam vsebuje integrale racionalnih funkcij.

$\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C\text{(za }n\ne -1\text{)}$ (Cavalierijev obrazec za kvadraturo)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Vsako racionalno funkcijo lahko integriramo tako, da uporabimo zgornje enačbe in delne ulomke z razvojem v vsoto funkcij z obliko:

$\frac{a}{(x-b{)}^n}$ in $\frac{ax+b}{{((x-c{)}^2+d^2)}^n}.$

Integrandi z obliko $\frac{x^m}{(a{x}^2+bx+c{)}^n}$

Za $a\ne 0:$

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

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

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

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

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

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

Integrandi z obliko $x^m{(a+b{x}^n)}^p$

• Integrand, ki ga dobimo ima enako obliko kot prvotni integrand tako, da lahko ponavljamo nižanje potenc tako, da nižamo potenci m in p proti nič.

• To zmanjšanje potenc lahko uporabimo za integrande, ki imajo celoštevilčne ali ulomljene potence.

$\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m+1}{(a+b{x}^n)}^p}{m+np+1}+\frac{anp}{m+np+1}\int x^m{(a+b{x}^n)}^{p-1}dx$

$\int x^m{(a+b{x}^n)}^{p}dx=-\frac{x^{m+1}{(a+b{x}^n)}^{p+1}}{an(p+1)}+\frac{m+n(p+1)+1}{an(p+1)}\int x^m{(a+b{x}^n)}^{p+1}dx$

$\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m+1}{(a+b{x}^n)}^p}{m+1}-\frac{bnp}{m+1}\int x^{m+n}{(a+b{x}^n)}^{p-1}dx$

$\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m-n+1}{(a+b{x}^n)}^{p+1}}{bn(p+1)}-\frac{m-n+1}{bn(p+1)}\int x^{m-n}{(a+b{x}^n)}^{p+1}dx$

$\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m-n+1}{(a+b{x}^n)}^{p+1}}{b(m+np+1)}-\frac{a(m-n+1)}{b(m+np+1)}\int x^{m-n}{(a+b{x}^n)}^{p}dx$

$\int x^m{(a+b{x}^n)}^{p}dx=\frac{x^{m+1}{(a+b{x}^n)}^{p+1}}{a(m+1)}-\frac{b(m+n(p+1)+1)}{a(m+1)}\int x^{m+n}{(a+b{x}^n)}^{p}dx$

Integrandi z obliko $x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^q$

• Nastali integrandi imajo enako obliko kot prvotni integrandi, to pa pomeni, da se zniževanje potence lahko ponavlja z zmanjševanjem potenc m, p in q proti 0.

• To zmanjševanje potenc se lahko uporabi za integrande, ki imajo celoštevilčne ali ulomljene eksponente.

• Posebni primer takšnega zmanjševanja potenc se lahko uporabi za integrande v obliki ${(a+b{x}^n)}^p{(c+d{x}^n)}^q$ in $x^m{(a+b{x}^n)}^p{(c+d{x}^n)}^q$ s postavitvijo m in/ali B na nič.

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=-\frac{(Ab-aB)x^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{abn(p+1)}+\frac{1}{abn(p+1)}\cdot$

$\int x^m(c(Abn(p+1)+(Ab-aB)(m+1))+d(Abn(p+1)+(Ab-aB)(m+nq+1))x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{B{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{b(m+n(p+q+1)+1)}+\frac{1}{b(m+n(p+q+1)+1)}\cdot$

$\int x^m(c((Ab-aB)(1+m)+Abn(1+p+q))+(d(Ab-aB)(1+m)+Bnq(bc-ad)+Abdn(1+p+q))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=-\frac{(Ab-aB)x^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{an(bc-ad)(p+1)}+\frac{1}{an(bc-ad)(p+1)}\cdot$

$\int x^m(c(Ab-aB)(m+1)+An(bc-ad)(p+1)+d(Ab-aB)(m+n(p+q+2)+1)x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{B{x}^{m-n+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{bd(m+n(p+q+1)+1)}-\frac{1}{bd(m+n(p+q+1)+1)}\cdot$

$\int x^{m-n}(aBc(m-n+1)+(aBd(m+nq+1)-b(-Bc(m+np+1)+Ad(m+n(p+q+1)+1)))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{A{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{ac(m+1)}+\frac{1}{ac(m+1)}\cdot$

$\int x^{m+n}(aBc(m+1)-A(bc+ad)(m+n+1)-An(bcp+adq)-Abd(m+n(p+q+2)+1)x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{A{x}^{m+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^q}{a(m+1)}-\frac{1}{a(m+1)}\cdot$

$\int x^{m+n}(c(Ab-aB)(m+1)+An(bc(p+1)+adq)+d((Ab-aB)(m+1)+Abn(p+q+1))x^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q-1}dx$

$\int x^m(A+B{x}^n){(a+b{x}^n)}^p{(c+d{x}^n)}^{q}dx=\frac{(Ab-aB)x^{m-n+1}{(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q+1}}{bn(bc-ad)(p+1)}-\frac{1}{bn(bc-ad)(p+1)}\cdot$

$\int x^{m-n}(c(Ab-aB)(m-n+1)+(d(Ab-aB)(m+nq+1)-bn(Bc-Ad)(p+1))x^n){(a+b{x}^n)}^{p+1}{(c+d{x}^n)}^{q}dx$

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