9. Partial Fractions
a2. General Decompositions
e. Non-Repeated Quadratic Factors
Find the partial fraction decomposition for \(\dfrac{13}{x^3-4x^2+13x}\).
We factor the denominator: \[ x^3-4x^2+13x=x(x^2-4x+13) \] The quadratic factor is irreducible because its discriminant is negative: \[ b^2-4ac=16-52=-36. \] So we complete the square on the quadratic term:
Completing the square is not necessary but it helps with finding the coefficients in the partial fraction decomposition and it helps with the integration.
\[ x^2-4x+13=(x^2-4x+4)+9=(x-2)^2+3^2 \] We now write the general partial fraction expansion and clear the denominator: \[ \dfrac{13}{x(x^2-4x+13)} =\dfrac{A}{x}+\dfrac{B(x-2)+C}{(x-2)^2+3^2} \] \[ 13=A\left((x-2)^2+3^2\right)+(B(x-2)+C)x \] To find the coefficient \(A\) for the linear factor, we plug in the root \(x=0\). \[ 13=A(13) \qquad \Longrightarrow \qquad A=1 \] To find the constant coefficient \(C\) for the quadratic term, we plug in \(x=2\): \[ 13=A(9)+(C)2=9+2C \qquad \Longrightarrow \qquad C=2 \] Finally to find the linear coefficient \(B\) for the quadratic term, we plug in any other number, say \(x=1\): \[ 13=A(10)+(-B+C)(1)=10-B+2 \qquad \Longrightarrow \qquad B=-1 \] So the partial fraction expansion is \[ \dfrac{13}{x^3-4x^2+13x}=\dfrac{1}{x}+\dfrac{-(x-2)+2}{(x-2)^2+3^2} \] Don't simplify! This is the form you will need for the quadratic term when you integrate.
We check by multiplying out the answer: \[\begin{aligned} \dfrac{1}{x}&+\dfrac{-(x-2)+2}{(x-2)^2+3^2} \\ &=\dfrac{(x-2)^2+3^2+x[-(x-2)+2]}{x((x-2)^2+3^2)} \\ &=\dfrac{13}{x^3-4x^2+13x} \end{aligned}\]
Completing the square in the quadratic term not only helps with the integration (as you will see later), but it also helps identify the number \(x=2\) needed for finding the coefficient \(C\).
Find the partial fraction expansion for \(\dfrac{1}{x^4-16}\).
\(\dfrac{1}{x^4-16} =\dfrac{1}{32}\left[-\dfrac{4}{x^2+4}-\dfrac{1}{x+2}+\dfrac{1}{x-2}\right]\)
To find the partial fraction expansion for \(\dfrac{1}{x^4-16}\), we first factor the denominator: \[ x^4-16=(x^2+4)(x^2-4)=(x^2+4)(x+2)(x-2) \] The general partial fraction decomposition is \[ \dfrac{1}{(x^2+4)(x+2)(x-2)} =\dfrac{Ax+B}{x^2+4}+\dfrac{C}{x+2}+\dfrac{D}{x-2} \] We clear the denominator: \[ 1=(Ax+B)(x+2)(x-2)+C(x^2+4)(x-2)+D(x^2+4)(x+2) \] To find the coefficients, we plug in \(x=2,-2,0,1\):
- \(x=2\): \[\begin{aligned} 1&=(2A+B)(4)(0)+C(8)(0)+D(8)(4)=32D \\ &\Longrightarrow \qquad D=\dfrac{1}{32} \end{aligned}\]
- \(x=-2\): \[\begin{aligned} 1&=(-2A+B)(0)(-4)+C(8)(-4)+D(8)(0)=-32C \\ &\Longrightarrow \qquad C=-\dfrac{1}{32} \end{aligned}\]
- \(x=0\): \[\begin{aligned} 1&=(B)(2)(-2)+C(4)(-2)+D(4)(2)=-4B-8C+8D=-4B+\dfrac{1}{2} \\ &\Longrightarrow \qquad B=-\dfrac{1}{8} \end{aligned}\]
- \(x=1\): \[\begin{aligned} 1&=(A+B)(3)(-1)+C(5)(-1)+D(5)(3)=-3A-3B-5C+15D \\ &=-3A+\dfrac{3}{8}+\dfrac{5}{32}+\dfrac{15}{32}=-3A+1 \\ &\Longrightarrow \qquad A=0 \end{aligned}\]
In summary the coefficients are \[ A=0 \qquad B=-\dfrac{1}{8} \qquad C=-\dfrac{1}{32} \qquad D=\dfrac{1}{32} \] So the partial fraction expansion is \[ \dfrac{1}{x^4-16} =\dfrac{1}{32}\left[-\dfrac{4}{x^2+4}-\dfrac{1}{x+2}+\dfrac{1}{x-2}\right] \]
We check by adding up the right side: \[\begin{aligned} \dfrac{1}{32}&\left[-\dfrac{4}{x^2+4}-\dfrac{1}{x+2}+\dfrac{1}{x-2}\right] \\ &=\dfrac{1}{32}\left[-\dfrac{4}{x^2+4}+\dfrac{4}{x^2-4}\right] \\ &=\dfrac{1}{32}\left[-\dfrac{-4(x^2-4)+4(x^2+4)}{x^4-16}\right] \\ &=\dfrac{1}{x^4-16} \end{aligned}\]