Differentials and integrals: Integration techniques
Partial fraction decomposition
For rational functions, an effective approach is sometimes to rewrite a rational function as a sum of simpler rational functions. This method is known as partial fraction decomposition. A few examples illustrate this approach.
Example 1 \[\int \frac{1}{y(1-y)}\,\dd y\]
In partial fraction decomposition one first tries to find \(A\) and \(B\) such that \[\frac{1}{y(1-y)}=\frac{A}{y}+\frac{B}{1-y}.\] In order to find \(A\) and \(B\), we work out the right-hand side by bringing the terms under a common denominator: \[\begin{aligned}\frac{A}{y}+\frac{B}{1-y}&=\frac{A(1-y)}{y(1-y)}+\frac{B y}{y(1-y)}\\ \\ &=\frac{A(1-y)+B y}{y(1-y)}\\ \\ &=\frac{(B-A)y+A}{y(1-y)}\end{aligned}\] Thus, we get the equations \(B-A=0\) and \(A=1\). So \(A=1\) and \(B=1\), that is, \[\frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}\] Then the following reasoning is correct: \[\begin{aligned}\int\frac{1}{y(1-y)}\,\dd y&=\int\frac{1}{y}\,\dd y+\int\frac{1}{1-y}\dd y\\ \\ &=\ln\bigl(|y|\bigr)-\ln\bigl(|1-y|\bigr)+c\end{aligned}\]
Not all denominators of rational functions can be split to some factors of degree 1 as above. Some polynomials have factors of degree 2 that cannot be further split into real factors. See the following example to find out how to use partial fraction decomposition in such case. We illustrate below how the method works.
Example 2 \[\int \frac{1}{x^3+4x}\,\dd x\]
In partial fraction decomposition one first tries to find \(A\) \(A\), \(B\) and \(C\) such that \[\begin{aligned}\frac{1}{x^3+4x} &= \frac{1}{x(x^2+4)}\\ \\ &=\frac{A}{x}+\frac{Bx+C}{x^2+4}, \end{aligned}\] In order to find \(A\), \(B\) and \(C\), we work out the right-hand side by bringing the terms under a common denominator: \[\begin{aligned}\frac{x}{x^3+4x} &=\frac{A(x^2+4) + (Bx+C)x}{x^3+4x}\\ \\&=\frac{(A+B)x^2 +Cx + 4A}{x^3+4x}\end{aligned}\] Thus, we get the equations \(A+B = 0\), \(C = 0\), and \(4A = 1\). So \(A=\frac{1}{4}\), \(B=-\frac{1}{4}\), \(C=0\). We get \[\begin{aligned} \int \frac{1}{x^3+4x}\,\dd x &=\frac{1}{4}\int\frac{1}{x}\dd x -\frac{1}{4} \int \frac{x}{x^2+4}\dd x\\ \\ &=\frac{1}{4}\ln|x| -\frac{1}{8}\ln(x^2+4)+c \end{aligned}\] In the second logarithmic term, we omit absolute value bars because \(x^2+4\) is always positive.
Another situation that can occur is that the denominator can be factorised to factors of degree 1 or 2 that occur more than once. See below for a worked example in that situation.
