Learning Lab

To express an algebraic fraction as partial fractions:

1. Factorize the denominator (using the guidelines above);

2. Write the algebraic fraction in partial fraction form with unknown constants \(A,A_{i},B\) and \(C\);

3. Add the partial fractions;

4. Equate coefficients, or substitute values of \(x\) to determine the value of the unknown constants.

Example 1: Find the integral of \(\frac{x-5}{x^{2}+2x-3}\) with respect to \(x\).

Solution: We first decompose the integrand into partial fractions. Since \(x^{2}+2x-3=\left(x+3\right)\left(x-1\right)\), we are dealing with the case where the denominator has two distinct linear factors. Therefore we factorize as follows: \[\begin{align*} \frac{x-5}{x^{2}+2x-3} & =\frac{x-5}{\left(x+3\right)\left(x-1\right)}\\ & =\frac{A}{x+3}+\frac{B}{x-1}. \end{align*}\] Now we have to determine the constants \(A\) and \(B.\) First add the partial fractions. We do this by finding a common denominator, in this case \(\left(x+3\right)\left(x-1\right):\) \[\begin{align*} \frac{A}{x+3}+\frac{B}{x-1} & =\frac{A}{\left(x+3\right)}\times\frac{\left(x-1\right)}{\left(x-1\right)}+\frac{B}{\left(x-1\right)}\times\frac{\left(x+3\right)}{\left(x+3\right)}\\ & =\frac{A\left(x-1\right)+B\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}\\ & =\frac{Ax-A+Bx+3B}{\left(x+3\right)\left(x-1\right)}\\ & =\frac{Ax+Bx-A+3B}{x^{2}+2x-3}\\ & =\frac{\left(A+B\right)x-A+3B}{x^{2}+2x-3} \end{align*}\] That is, \[\begin{align*} \frac{x-5}{x^{2}+2x-3} & =\frac{\overbrace{\left(A+B\right)}^{x-\textrm{coefficient}}x\overbrace{-A+3B}^{\textrm{constant term}}}{x^{2}+2x-3}. \end{align*}\] The constants \(A\) and \(B\) can be found by equating the coefficients of terms on the left and right sides of the equation. On the left, the coefficient of \(x\) is \(1\) and on the right the coefficient is \(A+B\) . Similarly the constant term on the left side is \(-5\) and on the right it is \(-A+3B\). This gives us the following simultaneous equations: \[\begin{align*} A+B & =1 & \left(1\right)\\ -A+3B & =-5 & \left(2\right) \end{align*}\] These equations are easily solved by elimination. Adding the equations we get \[\begin{align*} 4B & =-4\\ B & =-1. \end{align*}\] Substituting for \(B\) in equation \(\left(1\right)\) gives \[\begin{align*} A-1 & =1\\ A & =2. \end{align*}\] Finally we have: \[\begin{align*} \frac{x-5}{x^{2}+2x-3} & =\frac{A}{x+3}+\frac{B}{x-1}\\ & =\frac{2}{x+3}+\frac{-1}{x-1}\\ & =\frac{2}{x+3}-\frac{1}{x-1}. \end{align*}\] We can now evaluate the integral \[\begin{align*} \int\frac{x-5}{x^{2}+2x-3}dx & =\int\left(\frac{2}{x+3}-\frac{1}{x-1}\right)dx\\ & =\int\frac{2}{x+3}dx-\int\frac{1}{x-1}dx\\ & =2\ln\left|x+3\right|-\ln\left|x-1\right|+c\;\ \ \ \ \ ,c\textrm{ being a constant. } \end{align*}\] Using the laws of logarithms, \(a\ln\left(x\right)=\ln\left(x^{a}\right)\) and \(\ln\left(a\right)-\ln\left(b\right)=\ln\frac{a}{b}\) , we can also write the answer as: \[\begin{align*} \int\frac{x-5}{x^{2}+2x-3}dx & =2\ln\left|x+3\right|-\ln\left|x-1\right|+c\;\ \ \ \ \ \ c\textrm{ being a constant. }\\ & =\ln\left(x+3\right)^{2}-\ln\left|x-1\right|+c\\ & =\ln\frac{\left(x+3\right)^{2}}{\left|x-1\right|}+c \end{align*}\]

Example 2: Find the integral of \(\frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5}.\)

We first factorize the denominator: \[\begin{align*} x^{3}-5x^{2}+x-5 & =x^{2}\left(x-5\right)+x-5\\ & =\left(x^{2}+1\right)\left(x-5\right). \end{align*}\] In this case we have a quadratic factor and a linear factor. Therefore we factorize as follows: \[\begin{align*} \frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5} & =\frac{Ax+B}{x^{2}+1}+\frac{C}{x+5}\\ & =\frac{\left(Ax+B\right)\left(x-5\right)}{\left(x^{2}+1\right)\left(x-5\right)}+\frac{C\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x-5\right)}\\ & =\frac{Ax^{2}-5Ax+Bx-5B+Cx^{2}+C}{\left(x^{2}+1\right)\left(x+5\right)}\\ & =\left(A+C\right)x^{2}+\left(-5A+B\right)x-5B+C \end{align*}\] Equating coefficients of \(x^{2}\), \(x\) and the constant gives the three equations: \[\begin{align*} A+C & =1 & & \left(1\right)\\ -5A+B & =-3 & & \left(2\right)\\ -5B+C & =16. & & \left(3\right) \end{align*}\] which may be solved simultaneously as follows. Multiply \(\left(1\right)\) by \(5\): \[\begin{align*} 5A+5C & =5 & & \left(4\right) \end{align*}\] and add equation \(\left(2\right)\) to \(\left(4\right)\) to get: \[\begin{align*} B+5C & =-3+5\\ & =2\\ 5B+25C & =10\;\textrm{\;multiplying by $5$ .} & & \left(5\right) \end{align*}\] Adding equation \(\left(5\right)\) to \(\left(3\right)\) gives: \[\begin{align*} 26C & =26\\ C & =1. & & \left(6\right) \end{align*}\] Substituting \(\left(6\right)\) into \(\left(1\right)\) gives
\[\begin{align*} A+1 & =1\\ A & =0. & & \left(7\right) \end{align*}\] Finally substituting \(A=0\) in \(\left(2\right)\) gives \[\begin{align*} B & =-3. & & \left(8\right) \end{align*}\] Therefore the partial fraction decomposition is: \[\begin{align*} \frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5} & =\frac{Ax+B}{x^{2}+1}+\frac{C}{x-5}\\ & =\frac{-3}{x^{2}+1}+\frac{1}{x-5}. \end{align*}\]

We can now do the integral: \[\begin{align*} \int\left(\frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5}\right)dx & =\int\left(\frac{-3}{x^{2}+1}+\frac{1}{x-5}\right)dx\;\;\textrm{using partial fractions}\\ & =\int\frac{1}{x-5}dx-3\int\frac{1}{x^{2}+1}dx\\ & =\ln\left|x-5\right|-3\tan^{-1}\left(x\right)+c,\ \ \;c\textrm{ being a constant.} \end{align*}\]