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
: \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|
: \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
: \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
: \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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