Popis integrala racionalnih funkcija

Slijedi popis integrala (antiderivacija funkcija) racionalnih funkcija. Za potpun popis integrala funkcija, pogledati tablica integrala i popis integrala.

[math]\displaystyle{ \int (ax + b)^n dx }[/math]	[math]\displaystyle{ = \frac{(ax + b)^{n+1}}{a(n + 1)} + C \qquad\mbox{(za } n\neq -1\mbox{)}\,\! }[/math]
[math]\displaystyle{ \int\frac{1}{ax + b} dx }[/math]	[math]\displaystyle{ = \frac{1}{a}\ln\left|ax + b\right| + C }[/math]
[math]\displaystyle{ \int x(ax + b)^n dx }[/math]	[math]\displaystyle{ = \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1} + C \qquad\mbox{(za }n \not\in \{-1, -2\}\mbox{)} }[/math]

[math]\displaystyle{ \int\frac{x}{ax + b} dx }[/math]	[math]\displaystyle{ = \frac{x}{a} - \frac{b}{a^2}\ln\left|ax + b\right| + C }[/math]
[math]\displaystyle{ \int\frac{x}{(ax + b)^2} dx }[/math]	[math]\displaystyle{ = \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right| + C }[/math]
[math]\displaystyle{ \int\frac{x}{(ax + b)^n} dx }[/math]	[math]\displaystyle{ = \frac{a(1 - n)x - b}{a^2(n - 1)(n - 2)(ax + b)^{n-1}} + C \qquad\mbox{(za } n\not\in \{1, 2\}\mbox{)} }[/math]

[math]\displaystyle{ \int\frac{x^2}{ax + b} dx }[/math]	[math]\displaystyle{ = \frac{1}{a^3}\left(\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left|ax + b\right|\right) + C }[/math]
[math]\displaystyle{ \int\frac{x^2}{(ax + b)^2} dx }[/math]	[math]\displaystyle{ = \frac{1}{a^3}\left(ax + b - 2b\ln\left|ax + b\right| - \frac{b^2}{ax + b}\right) + C }[/math]
[math]\displaystyle{ \int\frac{x^2}{(ax + b)^3} dx }[/math]	[math]\displaystyle{ = \frac{1}{a^3}\left(\ln\left|ax + b\right| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right) + C }[/math]
[math]\displaystyle{ \int\frac{x^2}{(ax + b)^n} dx }[/math]	[math]\displaystyle{ = \frac{1}{a^3}\left(-\frac{(ax + b)^{3-n}}{(n-3)} + \frac{2b (a + b)^{2-n}}{(n-2)} - \frac{b^2 (ax + b)^{1-n}}{(n - 1)}\right) + C \qquad\mbox{(za } n\not\in \{1, 2, 3\}\mbox{)} }[/math]

[math]\displaystyle{ \int\frac{1}{x(ax + b)} dx }[/math]	[math]\displaystyle{ = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right| + C }[/math]
[math]\displaystyle{ \int\frac{1}{x^2(ax+b)} dx }[/math]	[math]\displaystyle{ = -\frac{1}{bx} + \frac{a}{b^2}\ln\left|\frac{ax+b}{x}\right| + C }[/math]
[math]\displaystyle{ \int\frac{1}{x^2(ax+b)^2} dx }[/math]	[math]\displaystyle{ = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x}\right|\right) + C }[/math]
[math]\displaystyle{ \int\frac{1}{x^2+a^2} dx }[/math]	[math]\displaystyle{ = \frac{1}{a}\arctan\frac{x}{a}\,\! + C }[/math]
[math]\displaystyle{ \int\frac{1}{x^2-a^2} dx = }[/math]	[math]\displaystyle{ \begin{cases} -\frac{1}{a}\,\mathrm{arctanh}\frac{x}{a} = \frac{1}{2a}\ln\frac{a-x}{a+x} + C \qquad\mbox{(za }|x| \lt |a|\mbox{)}\,\! \\ -\frac{1}{a}\,\mathrm{arccoth}\frac{x}{a} = \frac{1}{2a}\ln\frac{x-a}{x+a} + C \qquad\mbox{(za }|x| \gt |a|\mbox{)}\,\! \end{cases} }[/math]

[math]\displaystyle{ \int\frac{1}{ax^2+bx+c} dx = }[/math]

[math]\displaystyle{ \begin{cases} \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C \qquad\mbox{(za }4ac-b^2\gt 0\mbox{)} \\ -\frac{2}{\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} = \frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| + C \qquad\mbox{(za }4ac-b^2\lt 0\mbox{)} \\ -\frac{2}{2ax+b}+ C \qquad\mbox{(za }4ac-b^2=0\mbox{)} \end{cases} }[/math]

[math]\displaystyle{ \int\frac{x}{ax^2+bx+c} dx }[/math]

[math]\displaystyle{ = \frac{1}{2a}\ln\left|ax^2+bx+c\right|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c} + C }[/math]

[math]\displaystyle{ \int\frac{mx+n}{ax^2+bx+c} dx = }[/math]

[math]\displaystyle{ \begin{cases} \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C \qquad\mbox{(za }4ac-b^2\gt 0\mbox{)} \\ \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\mathrm{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C \qquad\mbox{(za }4ac-b^2\lt 0\mbox{)} \\ \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)} + C \,\,\,\,\,\,\,\,\,\, \qquad\mbox{(za }4ac-b^2=0\mbox{)} \end{cases} }[/math]

[math]\displaystyle{ \int\frac{1}{(ax^2+bx+c)^n} dx= \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}\int\frac{1}{(ax^2+bx+c)^{n-1}} dx\,\! + C }[/math]

[math]\displaystyle{ \int\frac{x}{(ax^2+bx+c)^n} dx= \frac{bx+2c}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int\frac{1}{(ax^2+bx+c)^{n-1}} dx\,\! + C }[/math]

[math]\displaystyle{ \int\frac{1}{x(ax^2+bx+c)} dx= \frac{1}{2c}\ln\left|\frac{x^2}{ax^2+bx+c}\right|-\frac{b}{2c}\int\frac{1}{ax^2+bx+c} dx + C }[/math]

Bilo koja racionalna funkcija se može integrirati rabeći gornje jednadžbe i parcijalne razlomke u integriranju, dekompozicijom racionalne funkcije u zbroj funkcija oblika:

[math]\displaystyle{ \frac{ex + f}{\left(ax^2+bx+c\right)^n} }[/math].