In the last two sections integrals involving trigonometric functions were simplified by using various trigonometric identities. For integrals of rational functions—quotients of polynomials—some algebraic identities (e.g. \(x^2 - a^2 = (x-a)(x+a)\)) will be useful in the method of partial fractions. The idea behind this method is simple: replace a complicated rational function with simpler ones that are easy to integrate.
For example, there is no formula for evaluating the integral
\[\int \frac{\dx}{x^2 + x} \nonumber \]
but notice that you can write
\[\frac{1}{x^2 + x} ~=~ \frac{1}{x\,(x+1)} ~=~ \frac{1}{x} ~-~ \frac{1}{x+1} ~, \nonumber \]
called the partial fraction decomposition of \(\frac{1}{x^2 + x}\), so that
\[\begin{aligned} \int \frac{\dx}{x^2 + x} ~&=~ \int \left(\frac{1}{x} ~-~ \frac{1}{x+1}\right)~\dx\
\[6pt] &=~ \ln\,\abs{x} ~-~ \ln\,\abs{x+1} ~+~ C ~.\end{aligned} \nonumber \]
There is a systematic way to find this decomposition. First, assume that
\[\frac{1}{x\,(x+1)} ~=~ \frac{A}{x} ~+~ \frac{B}{x+1} \nonumber \]
for some constants \(A\) and \(B\). Get a common denominator on the right side:
\[\begin{aligned} \frac{1}{x\,(x+1)} ~&=~ \frac{A\,(x+1) ~+~ Bx}{x\,(x+1)}\
\[4pt] \frac{0x ~+~ 1}{x\,(x+1)} ~&=~ \frac{(A + B)\,x ~+~ A}{x\,(x+1)}\end{aligned} \nonumber \]
Now equate coefficients in the numerators of both sides to solve for \(A\) and \(B\):
\[\begin{aligned} {3} \text{constant term}&: & A ~&=~ 1\\ \text{coefficient of $x$}&: \quad & A ~+~ B ~&=~ 0 \quad\Rightarrow\quad B ~=~ -A ~=~ -1\end{aligned} \nonumber \]
Thus,
\[\frac{1}{x\,(x+1)} ~=~ \frac{A}{x} ~+~ \frac{B}{x+1} ~=~ \frac{1}{x} ~+~ \frac{-1}{x+1} ~=~ \frac{1}{x} ~-~ \frac{1}{x+1} \nonumber \]
as before. The partial fraction method can be discussed in general, and its assumptions proved4, but only the simplest cases—linear and quadratic factors— will be considered here. In all cases it will be assumed that the degree of the polynomial in the numerator of the rational function is less than the degree of the polynomial in the denominator.
Start with the most basic case, similar to the example above:
Example \(\PageIndex{1}\): partfrac1
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Solution
Evaluate \(~\displaystyle\int \frac{\dx}{x^2 - 7x + 10}~\).
Solution: Since \(x^2 - 7x + 10 = (x-2)\,(x-5)\), then
\[\begin{aligned} \frac{1}{x^2 - 7x + 10} ~=~ \frac{1}{(x-2)\,(x-5)} ~&=~ \frac{A}{x-2} ~+~ \frac{B}{x-5}\
\[4pt] &=~ \frac{(A+B)\,x ~+~ (-5A - 2B)}{(x-2)\,(x-5)}\end{aligned} \nonumber \]
so that
\[\begin{aligned} {3} \text{coefficient of $x$}&: \quad & A ~+~ B ~&=~ 0 \quad\Rightarrow\quad B ~=~ -A\\ \text{constant term}&: & -5A ~-~ 2B ~&=~ 1 \quad\Rightarrow\quad -5A ~+~ 2A ~=~ 1 \quad\Rightarrow\quad A ~=~ -\frac{1}{3} ~~\text{and}~~ B ~=~ \frac{1}{3}\end{aligned} \nonumber \]
Thus,
\[\begin{aligned} \int \frac{\dx}{x^2 - 7x + 10} ~&=~ \int \left(\frac{-\frac{1}{3}}{x-2} ~+~ \frac{\frac{1}{3}}{x-5} \right)~\dx\
\[6pt] &=~ -\frac{1}{3}\,\ln\,\abs{x-2} ~+~ \frac{1}{3}\,\ln\,\abs{x-5} ~+~ C\end{aligned} \nonumber \]
If one linear factor in the denominator is repeated more than once, and all other factors are distinct, then use the following decomposition:
Example \(\PageIndex{1}\): partfrac2
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Solution
Evaluate \(~\displaystyle\int \frac{x^2 + x - 1}{x^3 + x^2}\;\dx~\).
Solution: Since \(x^3 + x^2 = x^2\,(x+1)\), then
\[\begin{aligned} \frac{x^2 + x - 1}{x^3 + x^2} ~=~ \frac{x^2 + x - 1}{x^2\,(x+1)} ~&=~ \frac{A}{x} ~+~ \frac{B}{x^2} ~+~ \frac{C}{x+1}\
\[4pt] &=~ \frac{Ax\,(x+1) ~+~ B\,(x+1) ~+~ Cx^2}{x^2\,(x+1)}\
\[4pt] &=~ \frac{(A+C)\,x^2 ~+~ (A+B)\,x ~+~ B}{x^2\,(x+1)}\end{aligned} \nonumber \]
so that
\[\begin{aligned} {3} \text{constant term}&: & B ~&=~ -1\\ \text{coefficient of $x$}&: \quad & A ~+~ B ~&=~ 1 \quad\Rightarrow\quad A ~=~ 1 ~-~ B ~~=~ 2\\ \text{coefficient of $x^2$}&: \quad & A ~+~ C ~&=~ 1 \quad\Rightarrow\quad C ~=~ 1 ~-~ A ~=~ -1\end{aligned} \nonumber \]
Thus,
\[\begin{aligned} \int \frac{x^2 + x - 1}{x^3 + x^2}\;\dx ~&=~ \int \left(\frac{2}{x} ~+~ \frac{-1}{x^2} ~+~ \frac{-1}{x+1} \right)~\dx\
\[6pt] &=~ 2\,\ln\,\abs{x} ~+~ \frac{1}{x} ~-~ \ln\,\abs{x+1} ~+~ C_0 \quad\text{($C_0 =~$ generic constant)}\end{aligned} \nonumber \]
Case 2 can be extended to more than one repeated factor—the partial fraction decomposition would then have more terms similar to those for the first repeated factor.
Example \(\PageIndex{1}\): partfrac3
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Solution
Evaluate \(~\displaystyle\int \frac{\dx}{x^2\,(x+1)^2}~\).
Solution: Expanding Case 2 to two repeated factors,
\[\begin{aligned} \frac{1}{x^2\,(x+1)^2} ~&=~ \frac{A}{x} ~+~ \frac{B}{x^2} ~+~ \frac{C}{x+1} ~+~ \frac{D}{(x+1)^2}\
\[4pt] &=~ \frac{Ax\,(x+1)^2 ~+~ B\,(x+1)^2 ~+~ Cx^2\,(x+1) ~+~ Dx^2}{x^2\,(x+1)^2}\
\[4pt] &=~ \frac{(A+C)\,x^3 ~+~ (2A+B+C+D)\,x^2 ~+~ (A+2B)\,x ~+~ B}{x^2\,(x+1)}\end{aligned} \nonumber \]
so that
\[\begin{aligned} {3} \text{constant term}&: & B ~&=~ 1\\ \text{coefficient of $x$}&: \quad & A ~+~ 2B ~&=~ 0 \quad\Rightarrow\quad A ~=~ -2B ~~=~ -2\\ \text{coefficient of $x^3$}&: \quad & A ~+~ C ~&=~ 0 \quad\Rightarrow\quad C ~=~ -A ~~=~ 2\\ \text{coefficient of $x^2$}&: \quad & 2A ~+~ B ~+~ C ~+~ D ~&=~ 0 \quad\Rightarrow\quad D ~=~ -2A ~-~ B ~-~ C ~=~ 1\end{aligned} \nonumber \]
Thus,
\[\begin{aligned} \int \frac{\dx}{x^2\,(x+1)^2} ~&=~ \int \left(\frac{-2}{x} ~+~ \frac{1}{x^2} ~+~ \frac{2}{x+1} ~+~ \frac{1}{(x+1)^2}\right)~\dx\
\[6pt] &=~ -2\,\ln\,\abs{x} ~-~ \frac{1}{x} ~+~ 2\,\ln\,\abs{x+1} ~-~ \frac{1}{x+1} ~+~ C_0 \quad\text{($C_0 =~$ generic constant)}\end{aligned} \nonumber \]
The partial fraction decompositions for quadratic factors are similar to those for linear factors, except the numerators in each partial fraction can now contain linear terms. A factor of the form \(ax^2 + bx + c\) is considered quadratic only if it cannot be factored into a product of linear terms (i.e. has no real roots) and \(a \ne 0\).
Example \(\PageIndex{1}\): partfrac4
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Solution
Evaluate \(~\displaystyle\int \frac{\dx}{(x^2+1)\,(x^2+4)}~\).
Solution: Neither \(x^2 + 1\) nor \(x^2 + 4\) has real roots, so by Case 3,
\[\begin{aligned} \frac{1}{(x^2+1)\,(x^2+4)} ~&=~ \frac{Ax+B}{x^2+1} ~+~ \frac{Cx+D}{x^2+4}\
\[4pt] &=~ \frac{(Ax+B)\,(x^2+4) ~+~ (Cx+D)\,(x^2+1)}{(x^2+1)\,(x^2+4)}\
\[4pt] &=~ \frac{(A+C)\,x^3 ~+~ (B+D)\,x^2 ~+~ (4A+C)\,x ~+~ (4B+D)}{(x^2+1)\,(x^2+4)}\end{aligned} \nonumber \]
so that
\[\begin{aligned} {3} \text{coefficient of $x^3$}&: \quad & A ~+~ C ~&=~ 0 \quad\Rightarrow\quad C ~=~ -A\\ \text{coefficient of $x^2$}&: \quad & B ~+~ D ~&=~ 0 \quad\Rightarrow\quad D ~=~ -B\\ \text{coefficient of $x$}&: \quad & 4A ~+~ C ~&=~ 0 \quad\Rightarrow\quad 4A ~-~ A ~=~ 0 \quad\Rightarrow\quad A ~=~ 0\\ \text{constant term}&: & 4B ~+~ D ~&=~ 1 \quad\Rightarrow\quad 4B ~-~ B ~=~ 1 \quad\Rightarrow\quad B ~=~ \frac{1}{3} ~~\text{and}~~ D ~=~ -\frac{1}{3}\end{aligned} \nonumber \]
Thus,
\[\begin{aligned} \int \frac{\dx}{(x^2+1)\,(x^2+4)} ~&=~ \int \left(\frac{\frac{1}{3}}{x^2+1} ~+~ \frac{-\frac{1}{3}}{x^2+4}\right)~\dx\
\[6pt] &=~ \frac{1}{3}\,\tan^{-1} x ~-~ \frac{1}{6}\,\tan^{-1}\left(\frac{x}{2}\right) ~+~ C_0\end{aligned} \nonumber \]
by formula ([eqn:atanint]) in Section 5.4.
A repeated quadratic factor is handled in the same way as a repeated linear factor:
Example \(\PageIndex{1}\): partfrac5
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Solution
Evaluate \(~\displaystyle\int \frac{\dx}{(x^2+1)^2\,(x^2+4)}~\).
Solution: Neither \(x^2 + 1\) nor \(x^2 + 4\) has real roots, and \(x^2 + 1\) is repeated, so by Case 4,
\[\begin{aligned} \frac{1}{(x^2+1)^2\,(x^2+4)} ~&=~ \frac{Ax+B}{x^2+1} ~+~ \frac{Cx+D}{(x^2+1)^2} ~+~ \frac{Ex+F}{x^2+4}\
\[4pt] &=~ \frac{(Ax+B)\,(x^2+1)\,(x^2+4) ~+~ (Cx+D)\,(x^2+4) ~+~ (Ex+F)\,(x^2+1)^2}{(x^2+1)\,(x^2+4)}\end{aligned} \nonumber \]
with the right side of the equation expanded as
\[\frac{(A+E)\,x^5 ~+~ (B+F)\,x^4 ~+~ (5A+C+2E)\,x^3 ~+~ (5B+D+2F)\,x^2 ~+~ (4A+4C+E)\,x ~+~ (4B+4D+F)}{(x^2+1)^2\,(x^2+4)} \nonumber \]
so that equating coefficients of both sides gives
\[\begin{aligned} {3} \text{coefficient of $x^5$}&: \quad & A \;+\; E \;&=\; 0 \quad\Rightarrow\quad E \;=\; -A\\ \text{coefficient of $x^4$}&: \quad & B \;+\; F \;&=\; 0 \quad\Rightarrow\quad F \;=\; -B\\ \text{coefficient of $x^3$}&: \quad & 5A \;+\; C \;+\; 2E \;&=\; 0 \quad\Rightarrow\quad 5A \;+\; C \;-\; 2A \;=\; 0 \quad\Rightarrow\quad C \;=\; -3A\\ \text{coefficient of $x^2$}&: \quad & 5B \;+\; D \;+\; 2F \;&=\; 0 \quad\Rightarrow\quad 5B \;+\; D \;-\; 2F \;=\; 0 \quad\Rightarrow\quad D \;=\; -3B\\ \text{coefficient of $x$}&: \quad & 4A \;+\; 4C \;+\; E \;&=\; 0 \quad\Rightarrow\quad 4A \;-\; 12A \;-\; A \;=\; 0 \enskip\Rightarrow\enskip A \;=\; 0 \enskip\Rightarrow\enskip C \;=\; 0 ~~\text{and}~~ E \;=\; 0\\ \text{constant term}&: \quad & 4B \;+\; 4D \;+\; F \;&=\; 1 \quad\Rightarrow\quad 4B \;-\; 12B \;-\; B \;=\; 1 \enskip\Rightarrow\enskip B \;=\; -\frac{1}{9} \enskip\Rightarrow\enskip D \;=\; \frac{1}{3} ~~\text{and}~~ F \;=\; \frac{1}{9}\end{aligned} \nonumber \]
Thus,
\[\begin{aligned} \int \frac{\dx}{(x^2+1)^2\,(x^2+4)} ~&=~ \int \left(\frac{-\frac{1}{9}}{x^2+1} ~+~ \frac{\frac{1}{3}}{(x^2+1)^2} ~+~ \frac{\frac{1}{9}}{x^2+4}\right)~\dx\
\[6pt] &=~ -\frac{1}{9}\,\tan^{-1} x ~+~ \frac{1}{3}\,\left(\frac{1}{2}\,\tan^{-1} x ~+~ \frac{x}{2\,(x^2 + 1)}\right) ~+~ \frac{1}{18}\,\tan^{-1}\left(\frac{x}{2}\right) ~+~ C_0\
\[6pt] &=~ \frac{1}{18}\,\tan^{-1} x ~+~ \frac{x}{6\,(x^2 + 1)} ~+~ \frac{1}{18}\,\tan^{-1}\left(\frac{x}{2}\right) ~+~ C_0\end{aligned} \nonumber \]
where the middle integral on the right is from Example
Example \(\PageIndex{1}\): trigsub2
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Solution
in Section 6.3.
When a rational function has a numerator with degree larger than its denominator, dividing the numerator by the denominator leaves the sum of a polynomial and a new rational function perhaps satisfying the conditions for Cases 1-4. When the numerator and denominator have the same degree, a trick like this might be easier.
\[\frac{x^2 + 2}{(x+1)\,(x+2)} ~=~ \frac{(x^2 + 3x + 2) - 3x}{(x+1)\,(x+2)} ~=~ \frac{x^2 + 3x + 2}{(x+1)\,(x+2)} ~-~ \frac{3x}{(x+1)\,(x+2)} ~=~ 1 ~-~ \frac{3x}{(x+1)\,(x+2)} \nonumber \]
The last rational function on the right can be integrated using partial fractions. [sec6dot4]
For Exercises 1-12, evaluate the given integral.
4
\(\displaystyle\int \frac{\dx}{x^2 - x}\)
\(\displaystyle\int \frac{x+1}{x^2 - x}\;\dx\)
\(\displaystyle\int \frac{\dx}{2x^2 + 3x - 2}\)
\(\displaystyle\int \frac{\dx}{x^2 + x - 6}\)
4
\(\displaystyle\int \frac{\dx}{x^4 - x^2}\vphantom{\displaystyle\int \frac{x^2}{(x-1)^2}}\)
\(\displaystyle\int \frac{x}{(x - 2)^3}\;\dx\vphantom{\displaystyle\int \frac{x^2}{(x-1)^2}}\)
\(\displaystyle\int \frac{x-1}{x^2\,(x+1)}\;\dx\vphantom{\displaystyle\int \frac{x^2}{(x-1)^2}}\)
\(\displaystyle\int \frac{x^2}{(x-1)^2}\;\dx\)
4
\(\displaystyle\int \frac{x-2}{x^2\,(x-1)^2}\;\dx\vphantom{\displaystyle\int \frac{(x-1)^2}{(x^2 + 1)^2}}\)
\(\displaystyle\int \frac{\dx}{x^4 - x^2}\vphantom{\displaystyle\int \frac{(x-1)^2}{(x^2 + 1)^2}}\)
\(\displaystyle\int \frac{\dx}{x^4 + 5x^2 + 4}\vphantom{\displaystyle\int \frac{(x-1)^2}{(x^2 + 1)^2}}\)
\(\displaystyle\int \frac{(x-1)^2}{(x^2 + 1)^2}\;\dx\)
[[1.]]
For all numbers \(a \ne b\) show that
\[\int \frac{\dx}{(x-a)\,(x-b)} ~=~ \frac{1}{a-b}\,\ln\;\ABS{\frac{x-a}{x-b}} ~+~ C ~. \nonumber \]
Let \(q(x) \;=\; (x-a_1)\,(x-a_2)\), where \(a_1 \ne a_2\). Show that
\[\frac{q'(x)}{q(x)} ~=~ \frac{1}{x-a_1} ~+~ \frac{1}{x-a_2} ~. \nonumber \]
For \(q(x)\) as in Exercise 14, show that
\[\frac{1}{q(x)} ~=~ \frac{1}{q'(a_1)\,(x-a_1)} ~+~ \frac{1}{q'(a_2)\,(x-a_2)} ~. \nonumber \]
Extend Exercise 15 to three distinct linear factors: if \(q(x) \;=\; (x-a_1)\,(x-a_2)\,(x-a_3)\,\) then
\[\frac{1}{q(x)} ~=~ \frac{1}{q'(a_1)\,(x-a_1)} ~+~ \frac{1}{q'(a_2)\,(x-a_2)} ~+~ \frac{1}{q'(a_3)\,(x-a_3)}~. \nonumber \]
This result can be extended to any \(n \ge 2\) distinct factors, though you do not need to prove that.
2
Find \(~\dfrac{d^{100}}{\dx^{100}}\;\left(\dfrac{1}{x^2 - 9x + 20}\right)\;\).
Find \(~\dfrac{d^{2020}}{\dx^{2020}}\;\left((x^2 - 1)^{-1}\right)\;\).
It is possible to use a form of partial fractions to evaluate integrals that are not rational functions. For example, evaluate the integral
\[\int \frac{\dx}{x + x^{4/3}} \nonumber \]
by finding constants \(A\), \(B\) and \(C\) such that
\[\frac{1}{x + x^{4/3}} ~=~ \frac{1}{x\,(1 + x^{1/3})} ~=~ \frac{A}{x} ~+~ \frac{B\,x^{-2/3} + C}{1 + x^{1/3}} ~. \nonumber \]
Notice how in the second partial fraction the highest power of \(x\) in the numerator is one less than in the denominator, similar to a partial fraction for a quadratic factor in a rational function. [[1.]]
Find a different way to evaluate the integral in Exercise 19.
