9. Partial Fractions

b2. Finding Coefficients - Quadratic

On the previous page, we discussed the techniques for finding the coefficients for terms with linear denominators in a partial fraction decomposition. On this page we discuss the techniques for terms with quadratic denominators.

Non-Repeated and Repeated Quadratic Denominators

As we discuss the techniques we will refer to two examples:

\[ \dfrac{x-12}{(x+1)\left((x-2)^2+4\right)} =\dfrac{A}{x+1}+\dfrac{B(x-2)+C}{(x-2)^2+4} \]

\[ \dfrac{x^3}{\left((x-2)^2+4\right)^2} =\dfrac{A(x-2)+B}{(x-2)^2+4}+\dfrac{C(x-2)+D}{\left((x-2)^2+4\right)^2} \]

The first has one non-repeated linear factor and one non-repeated quadratic factor in the denominator. The second has one repeated quadratic factor in the denominator.

As with linear factors, the first step is always:

For the two examples, we have:

\[ x-12=A\left((x-2)^2+4\right)+\left(B(x-2)+C\right)(x+1) \qquad \text{(1*)} \]

\[ x^3=\left(A(x-2)+B\right)\left((x-2)^2+4\right)+\left(C(x-2)+D\right) \qquad \text{(2*)} \]

The second step is to find the coefficients. There are several methods which may be mixed and matched:

This always works and gives all the coefficients, but it is not necessarily the most efficient method.

We expand each term in (1*) and collect terms with the same power of \(x\): \[\begin{aligned} x-12&=A(x^2-4x+8)+B(x^2-x-2)+C(x+1) \\ &=x^2(A+B)+x(-4A-B+C)+(8A-2B+C) \end{aligned}\] We equate the coefficients of each power of \(x\). This gives \(3\) equations in \(3\) unknowns: \[\begin{aligned} A+B&=0 \qquad (\text{1**}) \\ -4A-B+C&=1 \\ 8A-2B+C&=-12 \end{aligned}\] Equation (1**) says \(B=-A\). So the remaining equations become: \[\begin{aligned} -3A+C&=1 \\ 10A+C&=-12 \end{aligned}\] Subtracting the first equation from the second gives \(13A=-13\) or \(A=-1\). Substituting back gives \(B=1\) and \(C=-2\).

So the coefficients are: \[ A=-1 \qquad B=1 \qquad C=-2 \] and the partial fraction expansion is: \[ \dfrac{x-12}{(x+1)\left((x-2)^2+4\right)} =\dfrac{-1}{x+1}+\dfrac{(x-2)-2}{(x-2)^2+4} \]

We expand each term in (2*) and collect terms with the same power of \(x\): \[\begin{aligned} x^3&=A(x^3-6x^2+16x-16)+B(x^2-4x+8) \\ &\quad+C(x-2)+D \\ &=x^3(A)+x^2(-6A+B)+x(16A-4B+C) \\ &\quad+(-16A+8B-2C+D) \end{aligned}\] We equate the coefficients of each power of \(x\). This gives \(4\) equations in \(4\) unknowns: \[\begin{aligned} A&=1 \\ -6A+B&=0 \\ 16A-4B+C&=0 \\ -16A+8B-2C+D&=0 \end{aligned}\] Going down the equations, gives \(A=1\), \(B=6\), \(C=8\) and \(D=-16\).

So the coefficients are: \[ A=1 \qquad B=6 \qquad C=8 \qquad D=-16 \] and the partial fraction expansion is: \[ \dfrac{x^3}{\left((x-2)^2+4\right)^2} =\dfrac{(x-2)+6}{(x-2)^2+4}+\dfrac{8(x-2)-16}{\left((x-2)^2+4\right)^2} \]

Although the above method always works, it can be unwieldly if there are lots of coefficients. (A course on linear algebra teaches how to solve a system of linear equations with a large number of variables.) So it is useful to have a way to find some or all coefficients easily. Some of the techniques used for linear denominators can also help for terms with quadratic denominators.

For a non-repeated linear term \(\dfrac{A}{x-a}\) plugging in the obvious value \(x=a\) will give the coefficient \(A\). For a non-repeated quadratic term \(\dfrac{B(x-b)+C}{(x-b)^2+c^2}\) plugging in the obvious value \(x=b\) will eliminate the coefficient \(B\) from the equations and help find the coefficient \(C\). Then plugging in additional non-obvious values for \(x\) can help find the other coefficients.

We start with the equation with cleared denominators: \[ x-12=A\left((x-2)^2+4\right)+\left(B(x-2)+C\right)(x+1) \qquad \text{(1*)} \] Plugging in the obvious value \(x=-1\) will give the coefficient \(A\) for the linear term: \[ -13=A(13) \qquad \Longrightarrow \qquad A=-1 \] Plugging in the obvious value \(x=2\) and using the value of \(A\) will give the coefficient \(C\) for the quadratic term: \[ -10=A(4)+C(3)=-4+C(3) \qquad \Longrightarrow \qquad C=-2 \] Plugging in the non-obvious value \(x=0\) and using the values of \(A\) and \(C\) will give the remaining coefficient \(B\) for the quadratic term: \[ -12=A(8)+(B(-2)+C)=-8+B(-2)-2 \qquad \Longrightarrow \qquad B=1 \] These are of course the same coefficients as found before.

When there are repeated factors in the denominator, differentiating may give useful additional equations. After differentiating, we can again plug in obvious and non-obvious values for \(x\) and get equations for the coefficients. This can be repeated until all coefficients are found.

We start with the equation with cleared denominators: \[ x^3=\left(A(x-2)+B\right)\left((x-2)^2+4\right)+\left(C(x-2)+D\right) \qquad \text{(2*)} \] Taking the derivative gives: \[ 3x^2=A\left((x-2)^2+4\right)+\left(A(x-2)+B\right)\left(2(x-2)\right)+C \qquad \text{(2**)} \] Notice we used the product rule and did not simplify!
Plugging in the obvious value \(x=2\) into equations (2*) and (2**) gives expressions for the coefficients \(C\) and \(D\): \[ 8=B(4)+D \qquad \Longrightarrow \qquad D=8-4B \qquad \text{(2a)} \] \[ 12=A(4)+C \qquad \Longrightarrow \qquad C=12-4A \qquad \text{(2b)} \] Plugging in the non-obvious value \(x=0\) into equations (2*) and (2**) and using the formulas for \(C\) and \(D\), gives two equations for the coefficients \(A\) and \(B\): \[\begin{aligned} 0&=(A(-2)+B)(8)+(C(-2)+D) \\ &=-16A+8B-2C+D \\ &=-16A+8B-2(12-4A)+(8-4B) \\ &=-8A+4B-16 \qquad \Longrightarrow \qquad 2A-B=-4 \qquad \text{(2c)} \end{aligned}\] \[\begin{aligned} 0&=A(8)+(A(-2)+B)(-4)+C \\ &=16A-4B+C=16A-4B+(12-4A) \\ &=12A-4B+12 \qquad \Longrightarrow \qquad 3A-B=-3 \qquad \text{(2d)} \end{aligned}\] Subtracting (2c) from (2d) gives \(A=1\). Then substituting back into (2c) gives \(B=6\). Finally, substituting back into (2a) and (2b) gives \(C=8\) and \(D=-16\). These are of course the same coefficients as found before.

There is one further method that is very useful for terms with quadratic factors in the denominator.

Since this method involves complex numbers, it is beyond the scope of this class. It is discussed on this optional page. All problems in this class can be solved without using complex numbers, but this method could make some problems easier.

That completes the solution process for quadratic factors in the denominator. We are now ready to do the integrals.

You can also practice finding the Coefficients in a Partial Fraction Expansion by using the following Maplet (requires Maple on the computer where this is executed):

Partial Fractions: Finding Coefficients   Rate It

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