9. Partial Fractions

a2. General Decompositions

d. Repeated Linear Factors

Find the partial fraction expansion for \(\dfrac{x^2+1}{x^4+3x^3+3x^2+x}\).

We factor the denominator: \[ x^4+3x^3+3x^2+x=x(x^3+3x^2+3x+1)=x(x+1)^3 \] We write the generic partial fraction decomposition: \[ \dfrac{x^2+1}{x(x+1)^3} =\dfrac{A}{x}+\dfrac{B}{x+1}+\dfrac{C}{(x+1)^2}+\dfrac{D}{(x+1)^3} \] Notice that for the repeated factor \((x+1)^3\) we need to include denominators \((x+1)\), \((x+1)^2\) and \((x+1)^3\). Next we clear the denominator: \[ x^2+1=A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx \qquad (*) \] Don't simplify! Don't expand! We now try the evaluation trick that worked for non-repeating linear factors: Plug in \(x=0,-1\):

• \(x=0\): \[ 1=A(1)^3 \qquad \Longrightarrow \qquad A=1 \]
• \(x=-1\): \[ 2=D(-1) \qquad \Longrightarrow \qquad D=-2 \]

Notice that this only determined \(A\) which was the numerator for the non-repeating factor \(x\) and \(D\) which was the numerator for the highest power of the repeating factor \((x+1)\). There are \(3\) methods to find the remaining coefficients \(B\) and \(C\):

We can plug in for \(A\) and \(D\) in equation \((*)\), multiply out the right side, equate coefficients of powers of \(x\), and solve. This is tedious. So we don't recommend this method.

Since equation \((*)\) must be true for all \(x\), we just plug in two other random values of \(x\), say \(x=1,2\), as well as the known values of \(A\) and \(D\):

• \(x=1\): \[\begin{aligned} 2&=A(8)+B(1)(4)+C(1)(2)+D(1) \\ &=8A+4B+2C+D=8+4B+2C-2 \\ &\Longrightarrow \qquad 4B+2C=-4 \qquad \Longrightarrow \qquad 2B+C=-2 \end{aligned}\]
• \(x=2\): \[\begin{aligned} 5&=A(27)+B(2)(9)+C(2)(3)+D(2) \\ &=27A+18B+6C+2D=27+18B+6C-4 \\ &\Longrightarrow \qquad 18B+6C=-18 \qquad \Longrightarrow \qquad 3B+C=-3 \end{aligned}\]

Subtracting these equations gives \(B=-1\) and substituting back gives \(C=0\).

Since equation \((*)\) must be true for all \(x\), its derivative will also be true for all \(x\). So we differentiate \((*)\) and plug in \(x=-1\); then differentiate again and plug in \(x=-1\):

• We compute the first derivative and plug in \(A\) and \(D\): \[\begin{aligned} 2x&=A3(x+1)^2+B\left[(x+1)^2+x2(x+1)\right]+C(2x+1)+D \\ &=3(x+1)^2+B\left[(x+1)^2+x2(x+1)\right]+C(2x+1)-2 \end{aligned}\] At \(x=-1\): \[ -2=C(-1)-2 \qquad \Longrightarrow \qquad C=0 \]
• We compute the second derivative and plug in \(A\) and \(D\): \[\begin{aligned} 2&=A6(x+1)+B[2(x+1)+2(x+1)+x2]+C(2) \\ &=6(x+1)+B[2(x+1)+2(x+1)+x2] \end{aligned}\] At \(x=-1\): \[ 2=B(-2) \qquad \Longrightarrow \qquad B=-1 \]

Notice that as we take the derivatives, we use the product rule and chain rule rather than multiplying out first. This makes it easier to plug in the zeros of the denominator.

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In any case the the coefficients are \[ A=1\qquad B=-1\qquad C=0\qquad D=-2 \] and the partial fraction expansion is: \[ \dfrac{x^2+1}{x^4+3x^3+3x^2+x}=\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{2}{(x+1)^3} \]

Notice that some coefficients may be zero but there is no way to know that in advance.

We check by adding up the right side: \[\begin{aligned} \dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{2}{(x+1)^3} &=\dfrac{(x+1)^3-x(x+1)^2-2x}{x(x+1)^3} \\ &=\dfrac{x^2+1}{x^4+3x^3+3x^2+x} \end{aligned}\]