以下是部份有理函數的积分表。

 \int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n + 1)} +C \qquad (n\neq -1) 
\int\frac{1}{ax + b} dx= \frac{1}{a}\ln\left|ax + b\right| +C 
\int x(ax + b)^n dx = \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1} +C \qquad(n \not\in \{1, 2\})
\int\frac{x}{ax + b}dx = \frac{x}{a} - \frac{b}{a^2}\ln\left|ax + b\right| +C 
\int\frac{x}{(ax + b)^2}dx = \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right| +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 \qquad (n\not\in \{1, 2\} )
\int\frac{x^2}{ax + b}dx = \frac{1}{a^3}\left[\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left|ax + b\right|\right] +C 
\int\frac{x^2}{(ax + b)^2}dx = \frac{1}{a^3}\left(ax + b - 2b\ln\left|ax + b\right| - \frac{b^2}{ax + b}\right) +C 
\int\frac{x^2}{(ax + b)^3}dx = \frac{1}{a^3}\left[\ln\left|ax + b\right| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right] +C 
\int\frac{x^2}{(ax + b)^n}dx = \frac{1}{a^3}\left[-\frac{1}{(n- 3)(ax + b)^{n-3}} + \frac{2b}{(n-2)(ax + b)^{n-2}} - \frac{b^2}{(n - 1)(ax + b)^{n-1}}\right] +C \qquad
(n\not\in \{1, 2, 3\} )
\int\frac{dx}{x(ax + b)} =-\frac{1}{b}\ln\left|\frac{ax+b}{x}\right| +C 
\int\frac{dx}{x^2(ax+b)} = -\frac{1}{bx} + \frac{a}{b^2}\ln\left|\frac{ax+b}{x}\right| +C 
\int\frac{dx}{x^2(ax+b)^2} = -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 
\int\frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\frac{x}{a} +C 
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {artanh} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {a-x}{a+x}}+C\qquad {\mbox{(}}|x|<|a|{\mbox{)}}\,\!}
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}=-{\frac {1}{a}}\,\mathrm {arcoth} {\frac {x}{a}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C\qquad {\mbox{(}}|x|>|a|{\mbox{)}}\,\!}
\int\frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} +C ,(4ac-b^2>0)
\int\frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{b^2-4ac}}\,\mathrm{artanh}\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 \qquad\mbox{(}4ac-b^2<0\mbox{)}
\int\frac{dx}{ax^2+bx+c} = -\frac{2}{2ax+b} +C \qquad\mbox{(}4ac-b^2=0\mbox{)}
\int\frac{x}{ax^2+bx+c}dx = \frac{1}{2a}\ln\left|ax^2+bx+c\right|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c} +C 
\int\frac{mx+n}{ax^2+bx+c}dx = \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{(}4ac-b^2>0\mbox{)}
\int\frac{mx+n}{ax^2+bx+c}dx = \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{(}4ac-b^2<0\mbox{)}
\int\frac{mx+n}{ax^2+bx+c}dx = \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)} +C \qquad\mbox{(}4ac-b^2=0\mbox{)}
\int\frac{dx}{(ax^2+bx+c)^n} = \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{dx}{(ax^2+bx+c)^{n-1}} +C \,\!
\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{dx}{(ax^2+bx+c)^{n-1}} +C \,\!
對於任意的有理函數，我們都能通過部分分式(partial fraction)把該函數分拆為數個函數的總和，其中每個函數符合以下的形式：\frac{px + q}{\left(ax^2+bx+c\right)^n}。我們繼而能把每一個該種形式的函數作积分運算：

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

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积分表 |
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* 有理函數
* 無理函數
* 三角函數
* 反三角函數
* 雙曲函數
* 反雙曲函數
* 指數函數
* 對數函數
* 高斯函數

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