# antiderivative of rational function

The antiderivative of a rational function is always expressible in a closed form, which only can comprise, except a rational expression summand, summands of logarithms and arcustangents of rational functions.

One can justify the theorem by using the general form of the (unique) partial fraction decomposition

 $\displaystyle R(x)\;=\quad H(x)+$ $\displaystyle\sum_{i=1}^{m}\left(\frac{A_{i1}}{x\!-\!a_{i}}+\frac{A_{i2}}{(x\!% -\!a_{i})^{2}}+\ldots+\frac{A_{i\mu_{i}}}{(x\!-\!a_{i})^{\mu_{i}}}\right)$ $\displaystyle+$ $\displaystyle\sum_{j=1}^{n}\left(\frac{B_{j1}x\!+\!C_{j1}}{x^{2}\!+\!2p_{j}x\!% +\!q_{j}}+\frac{B_{j2}x\!+\!C_{j2}}{(x^{2}\!+\!2p_{j}x\!+\!q_{j})^{2}}+\ldots+% \frac{B_{j\nu_{j}}x\!+\!C_{j\nu_{j}}}{(x^{2}\!+\!2p_{j}x\!+\!q_{j})^{\nu_{j}}}% \right)\!,$

of the rational function $R(x)$ ; here, $H(x)$ is a polynomial    , the first sum expression is determined by the real zeroes $a_{i}$ of the denominator of $R(x)$, the second sum is determined by the real quadratic prime factors  $x^{2}\!+\!2p_{j}x\!+\!q_{j}$ of the denominator (which have no real zeroes).

The addends of the form $\displaystyle\frac{A}{(x\!-\!a)^{r}}$ in the first sum are integrated directly, giving

 $\displaystyle\int\!\frac{A}{x\!-\!a}\,dx\;=\;A\ln|x\!-\!a|+\mbox{constant}% \qquad(r=1)$ (1)

and

 $\displaystyle\int\!\frac{A}{(x\!-\!a)^{r}}\,dx\;=\;-\frac{A}{r\!-\!1}\!\cdot\!% \frac{1}{(x\!-\!a)^{r-1}}+\mbox{constant}\qquad(r>1).$ (2)

The remaining partial fractions  are of the form $\displaystyle\frac{Bx\!+\!C}{(x^{2}\!+\!2px\!+\!q)^{s}}$ where  $p^{2}  and $s$ is a positive integer.  Now we may write

 $x^{2}\!+\!2px\!+\!q\;=\;(x\!+\!p)^{2}\!+\!q\!-\!p^{2}\;=\;(q\!-\!p^{2})\!\left% [1+\left(\!\frac{x\!+\!p}{\sqrt{q\!-\!p^{2}}}\!\right)^{\!2}\right]$

and make the substitution

 $\displaystyle\frac{x\!+\!p}{\sqrt{q\!-\!p^{2}}}\;=\;t,$ (3)

i.e.  $x\,=\,t\sqrt{q\!-\!p^{2}}\!-\!p$,  getting

 $\displaystyle\int\!\frac{Bx\!+\!C}{(x^{2}\!+\!2px\!+\!q)^{s}}\,dx\;=\;\int\!% \frac{Et\!+\!F}{(1\!+\!t^{2})^{s}}\,dt\;=\;E\!\int\!\frac{t\,dt}{(1\!+\!t^{2})% ^{s}}+F\!\int\!\frac{dt}{(1\!+\!t^{2})^{s}}$ (4)

where $E$ and $F$ are certain constants.  In the case  $s=1$  we have

 $\displaystyle\int\!\frac{t\,dt}{1\!+\!t^{2}}\;=\;\frac{1}{2}\ln(1\!+\!t^{2})+% \mbox{constant}$ (5)

and in the case  $s>1$

 $\displaystyle\int\!\frac{t\,dt}{(1\!+\!t^{2})^{s}}\;=\;-\frac{1}{2(s\!-\!1)}\!% \cdot\!\frac{1}{(1\!+\!t^{2})^{s-1}}+\mbox{constant}.$ (6)

The latter addend of the right hand side of (4) is for  $s=1$  got from

 $\displaystyle\int\!\frac{dt}{1\!+\!t^{2}}\;=\;\arctan{t}+\mbox{constant}$ (7)

and for the cases $s>1$ on may first write

 $\int\!\frac{dt}{(1\!+\!t^{2})^{s}}\;=\;\int\!\frac{(1\!+\!t^{2})\!-\!t^{2}}{(1% \!+\!t^{2})^{s}}\,dt\;=\;\int\!\frac{dt}{(1\!+\!t^{2})^{s-1}}-\int\!t\!\cdot\!% \frac{t\,dt}{(1\!+\!t^{2})^{s}}.$

Using integration by parts in the last integral, this equation can be converted into the reduction formula

 $\displaystyle\int\!\frac{dt}{(1\!+\!t^{2})^{s}}\;=\;\frac{1}{2s\!-\!2}\!\cdot% \!\frac{t}{(1\!+\!t^{2})^{s-1}}+\frac{2n\!-\!3}{2n\!-\!2}\int\!\frac{dt}{(1\!+% \!t^{2})^{s-1}}.$ (8)

The assertion of the theorem follows from (1), …, (8).

Example.

 $\int\!\frac{dx}{1\!+\!x^{4}}\;=\;\frac{1}{4\sqrt{2}}\ln\frac{1\!+\!x\sqrt{2}\!% +\!x^{2}}{1\!-\!x\sqrt{2}\!+\!x^{2}}+\frac{1}{2\sqrt{2}}\arctan\frac{x\sqrt{2}% }{1\!-\!x^{2}}+C$
Title antiderivative of rational function AntiderivativeOfRationalFunction 2013-03-22 19:21:38 2013-03-22 19:21:38 pahio (2872) pahio (2872) 19 pahio (2872) Theorem msc 26A36 integration of rational functions IntegrationTechniques