9. Partial Fractions
a2. General Decompositions
f. Repeated Quadratic Factors
Find the partial fraction decomposition for \(\dfrac{100}{(x-1)\left((x-2)^2+3^2\right)^2}\).
The denominator is already factored and the square has already been completed in the quadratic term. So we write the generic partial fraction expansion: \[ \dfrac{100}{(x-1)\left((x-2)^2+3^2\right)^2} =\dfrac{A}{x-1}+\dfrac{B(x-2)+C}{(x-2)^2+3^2} +\dfrac{D(x-2)+E}{\left((x-2)^2+3^2\right)^2} \] Notice that we include a quadratic denominator with each power up to the power which appears in the original denominator. Next we clear the denominators: \[\begin{aligned} 100=&A\left((x-2)^2+3^2\right)^2+(B(x-2)+C)(x-1)\left((x-2)^2+3^2\right) \\ &+(D(x-2)+E)(x-1) \end{aligned}\] To find the coefficients we plug in numbers. The two obvious numbers are \(x=1,2\). We also use \(x=0,-1,-2\):
- \(x=1\): \[ 100=A(10)^2 \qquad \Longrightarrow \qquad A=1 \] From here on, we use this in the other equations.
- \(x=2\): \[\begin{aligned} 100&=A(9)^2+(C)(1)(9)+(E)(1)=81+9C+E \\ &\Longrightarrow \qquad 9C+E=9 \end{aligned}\]
- \(x=0\): \[\begin{aligned} 100&=A(13)^2+(-2B+C)(-1)(13)+(-2D+E)(-1) \\ &=169+26B-13C+2D-E \\ &\Longrightarrow \qquad 26B-13C+2D-E=-69 \end{aligned}\]
- \(x=-1\): \[\begin{aligned} 100&=A(18)^2+(-3B+C)(-2)(18)+(-3D+E)(-2) \\ &=324+108B-36C+6D-2E \\ &\Longrightarrow \qquad 108B-36C+6D-2E=-224 \end{aligned}\]
- \(x=-2\): \[\begin{aligned} 100&=A(25)^2+(-4B+C)(-3)(25)+(-4D+E)(-3) \\ &=625+300B-75C+12D-3E \\ &\Longrightarrow \qquad 300B-75C+12D-3E=-525 \end{aligned}\]
This gives \(4\) equations for the \(4\) unknowns \(B\), \(C\), \(D\) and \(E\). \[\begin{aligned} 9C+E &=19\qquad\qquad\qquad \text{(*)} \\ 26B-13C+2D-E &=-69 \\ 54B-18C+3D-E &=-112 \\ 100B-25C+4D-E &=-175 \end{aligned}\] We solve (*) for \(E=19-9C\) and substitute into the other \(3\) equations. \[\begin{aligned} 26B-4C+2D &=-50 \\ 54B-9C+3D &=-93 \\ 100B-16C+4D &=-156 \end{aligned}\] These simplify to \[\begin{aligned} 13B-2C+D &=-25\qquad\qquad \text{(**)} \\ 18B-3C+D &=-31 \\ 25B-4C+D &=-39 \end{aligned}\] We solve (**) for \(D=-25-13B+2C\) and substitute into the other \(2\) equations. \[\begin{aligned} 5B-C &=-6\qquad\qquad \text{(***)} \\ 12B-2C &=-14 \end{aligned}\] We solve (***) for \(C=5B+6\) and substitute into the other equation. \[ B=-1 \] Substituting back we find the coefficients are: \[ A=1\qquad B=-1\qquad C=1\qquad D=-10\qquad E=10 \] So the partial fraction expansion is \[ \dfrac{100}{(x-1)\left((x-2)^2+3^2\right)^2} =\dfrac{1}{x-1}+\dfrac{-(x-2)+1}{(x-2)^2+3^2} +\dfrac{-10(x-2)+10}{\left((x-2)^2+3^2\right)^2} \] Don't simplify! This is the form you will need for the quadratic terms when you integrate.
Find the partial fraction expansion for \(\dfrac{1}{x(x^2+1)^2}\).
Expand the right hand side instead of plugging in numbers.
\(\dfrac{1}{x(x^2+1)^2} =\dfrac{1}{x}-\dfrac{x}{x^2+1}-\dfrac{x}{(x^2+1)^2}\)
The generic expansion is \[ \dfrac{1}{x(x^2+1)^2} =\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2} \] We clear the denominator: \[ 1=A(x^2+1)^2+(Bx+C)x(x^2+1)+(Dx+E)x \] This time we expand the right hand side and collect terms: \[\begin{aligned} 1&=Ax^4+2Ax^2+A+Bx^4+Bx^2+Cx^3+xC+Dx^2+xE \\ &=(A+B)x^4+Cx^3+(2A+B+D)x^2+(C+E)x+A \end{aligned}\] Next we equate coefficients of each power of \(x\): \[ A+B=0 \qquad C=0 \qquad 2A+B+D=0 \qquad C+E=0 \qquad A=1 \] So the coefficients are: \[ A=1 \qquad B=-1 \qquad C=0 \qquad D=-1 \qquad E=0 \] and the partial fraction expansion is \[ \dfrac{1}{x(x^2+1)^2} =\dfrac{1}{x}-\dfrac{x}{x^2+1}-\dfrac{x}{(x^2+1)^2} \]
We check by adding up the right side: \[\begin{aligned} \dfrac{1}{x}-\dfrac{x}{x^2+1}-\dfrac{x}{(x^2+1)^2} &=\dfrac{(x^2+1)^2-x^2(x^2+1)-x^2}{x(x^2+1)^2} \\ &=\dfrac{1}{x(x^2+1)^2} \end{aligned}\]