Partial Fractions Done Right

The Case of Repeated Complex Roots

We wish to compute the partial partial fraction expansion for the irreducible fraction \[ \frac{p(x)}{(x^2 + ax + b)^r\phi(x)} \] with $r>1$, where the quadratic polynomial $x^2 + ax + b$ has complex roots, that is, $a^2 - 4b < 0$, and where $\phi(x)$ has no factor of $x^2 + ax + b$. The partial partial expansion has the form \[ \frac{p(x)}{(x^2 + ax + b)^r\phi(x)} = \frac{A_1 x + B_1}{x^2 + ax + b} + \frac{A_2 x + B_2}{(x^2 + ax + b)^2} + \cdots + \frac{A_r x + B_r}{(x^2 + ax + b)^r} + \frac{q(x)}{\phi(x)}. \]

The new idea is to introduce the auxiliary fraction \[ \frac{p(x)}{(x^2 + ax + t)\phi(x)} \] obtained by dropping the exponent $r$ and replacing the constant $b$ by a variable $t$. The auxiliary fraction admits a partial partial expansion of the form \[ \frac{p(x)}{(x^2+ax+t) \phi(x)} = \frac{A(t)x + B(t)}{x^2+ax+t} + \frac{q(x,t)}{\phi(x)}, \] where the coefficients $A(t)$ and $B(t)$ may be computed by the method described in The case of a simple complex root. Differentiating this with respect to $t$ we see that \[ -\frac{p(x)}{(x^2+ax+t)^2 \phi(x)} = \frac{A'(t)x + B'(t)}{x^2+ax+t} - \frac{A(t)x + B(t)}{(x^2+ax+t)^2} + \frac{q_t(x,t)}{\phi(x)}, \] where $q_t = \partial q/\partial t$.

To finish, we substitute $t=b$, and multiply both sides by $-1$ to get rid of the negative sign on the left hand side: \[ \frac{p(x)}{(x^2+ax+b)^2 \phi(x)} = -\frac{A'(b)x + B'(b)}{x^2+ax+b} + \frac{A(b)x + B(b)}{(x^2+ax+b)^2} - \frac{q_t(x,b)}{\phi(x)}. \] This is the desired partial partial fraction expansion for $r=2$. Further differentiations produce expressions for larger $r$.

In calculus we learn the differentiation rule for fractions: \[ \Big(\frac{u}{v} \Big)' = \frac{u'v - uv'}{v^2}. \] Alternatively, we may think of the ratio $\ds\frac{u}{v}$ as a product, $\ds u \cdot \frac{1}{v} $, then differentiate it by applying the product rule: \[ \Big(\frac{u}{v} \Big)' = \Big( u \cdot \frac{1}{v} \Big)' = u' \cdot \frac{1}{v} + u \cdot \Big( \frac{1}{v} \Big)' = u' \cdot \frac{1}{v} - u \cdot \frac{v'}{v^2} . \] The result is no different from the standard rule for fractions, but the way of thinking is. This alternative way of thinking is more suitable for computing the $t$ derivatives needed above.
Compute the partial fraction expansion of \[ \frac{3x^2+2x+1}{(x-1)(x^2+x+1)^2}. \]
Introduce the auxiliary fraction \[ \frac{3x^2+2x+1}{(x-1)(x^2+x+t)} \] and expand it into partial fractions: \begin{equation} \label{example1} \frac{3x^2+2x+1}{(x-1)(x^2+x+t)} = \frac{Ax+B}{x^2+x+t} + \frac{q(t)}{x-1}. \end{equation} We determine $A$ and $B$ following the procedure in The case of a simple complex root. Thus, we clear the identity \eqref{example1} of fractions \[ 3x^2+2x+1 = (x-1)(Ax+B) + (x^2+x+t)q(t) \] and expand: \[ 3x^2+2x+1 = Ax^2 + (B-A) x - B + (x^2+x+t)q(t), \] then replace $x^2$ by $-x-t$: \[ 3(-x-t)+2x+1 = A(-x-t) + (B-A) x - B \] and collect terms involving $x$: \[ -x-3t+1 = (B - 2A) x - tA-B. \] Equating the like powers of $x$ we arrive at \[ B-2A = -1, \qquad -tA - B = -3t+1, \] whence \[ A = \frac{3t}{t+2}, \qquad B = \frac{5t-2}{t+2}. \]
In preparation for differentiating $A$ and $B$, it is extremely helpful to reduce them to as simple forms as possible, otherwise the subsequent calculations can become too messy.
In the current problem it suffices to reduce the expressions for $A$ and $B$ to proper fraction: \begin{align*} A &= \frac{3t}{t+2} = \frac{3(t+2-2)}{t+2} = \frac{3(t+2)-6}{t+2} = 3 - \frac{6}{t+2}, \\ B &= \frac{5t-2}{t+2} = \frac{5(t+2-2)-2}{t+2} = \frac{5(t+2)-12}{t+2} = 5 - \frac{12}{t+2}. \end{align*} Then we compute the derivatives: \[ A' = \frac{6}{(t+2)^2}, \qquad B' = \frac{12}{(t+2)^2}. \] In particular, at $t=1$ we get: \[ A = 1, \quad B = 1, \quad A' = 2/3, \quad B' = 4/3. \]

Going back to \eqref{example1}, we differentiate both sides with respect to $t$: \begin{equation} \label{example2} -\frac{3x^2+2x+1}{(x-1)(x^2+x+t)^2} = \frac{A'x+B'}{x^2+x+t} - \frac{Ax+B}{(x^2+x+t)^2} + \frac{q'(t)}{x-1}, \end{equation} then substitute $t=1$: \[ -\frac{3x^2+2x+1}{(x-1)(x^2+x+1)^2} = \frac{(2/3)x+(4/3)}{x^2+x+1} - \frac{x+1}{(x^2+x+1)^2} + \frac{q'(1)}{x-1}, \] and multiply through by $-1$: \[ \frac{3x^2+2x+1}{(x-1)(x^2+x+1)^2} = -\frac{(2/3)x+(4/3)}{x^2+x+1} + \frac{x+1}{(x^2+x+1)^2} - \frac{q'(1)}{x-1}, \] The coefficient of the fraction $1/(x-1)$ on the right hand side may be computed by the method of The case of a simple real root as: \[ \frac{3x^2+2x+1}{\bbox[#ffdddd]{\color{#999999}{(x-1)}}(x^2+x+1)^2}\bigg|_{x=1} = \frac{6}{3^2} = \frac{2}{3}, \] therefore the full expansion is: \[ \frac{3x^2+2x+1}{(x-1)(x^2+x+1)^2} = -\frac{2(x+2)}{3(x^2+x+1)} + \frac{x+1}{(x^2+x+1)^2} + \frac{2}{3(x-1)}. \]

Compute the partial fraction expansion of \[ \frac{3x^2+2x+1}{(x-1)(x^2+x+1)^3}. \]
The auxiliary fraction is identical to that of the previous example, therefore $A(t)$ and $B(t)$ are as computed before. All we need is to compute the second derivatives: \[ A'' = -\frac{12}{(t+2)^3}, \quad B'' = -\frac{24}{(t+2)^3}, \] whence at $t=1$: \[ A'' = -\frac{4}{9} \quad B'' = -\frac{8}{9}. \] Now we differentiate \eqref{example2} with respect to $t$: \[ \frac{2(3x^2+2x+1)}{(x-1)(x^2+x+t)^3} = \frac{A''(t)x+B''(t)}{x^2+x+t} - \frac{2\big(A'(t)x+B'(t)\big)}{(x^2+x+t)^2} + \frac{2\big(A(t)x+B(t)\big)}{(x^2+x+t)^3} + \frac{q''(t)}{x-1}, \] and substitute $t=1$: \[ \frac{2(3x^2+2x+1)}{(x-1)(x^2+x+1)^3} = \frac{(-4/9)x+(-8/9)}{x^2+x+1} - \frac{2\big((2/3)x+(4/3)\big)}{(x^2+x+1)^2} + \frac{2(x+1)}{(x^2+x+1)^3} + \frac{q''(1)}{x-1}, \] and divide through by 2: \[ \frac{3x^2 + 2x + 1}{(x-1)(x^2 + x + 1)^3} = - \frac{2(x + 2)}{9(x^2 + x + 1)} - \frac{2(x + 2)}{3(x^2 + x + 1)^2} + \frac{x + 1}{(x^2 + x + 1)^3} + \frac{q''(1)/2}{(x-1)}. \] Finally, compute the value of the coefficient of the $1/(x-1)$ term on the right hand side: \[ \frac{3x^2+2x+1}{(x^2+x+1)^3}\bigg|_{x=1} = \frac{6}{3^3} = \frac{2}{9}, \] therefore the full expansion is: \[ \frac{3x^2 + 2x + 1}{(x-1)(x^2 + x + 1)^3} = \frac{x + 1}{(x^2 + x + 1)^3} - \frac{2(x + 2)}{3(x^2 + x + 1)^2} - \frac{2(x + 2)}{9(x^2 + x + 1)} + \frac{2}{9(x-1)}. \]
Compute the partial fraction expansion of \[ \frac{1}{(x^2+1)^2(x^2+9)^2}. \]
First we focus on the terms resulting from the $(x^2+1)^2$ factor. Later will will repeat the procedure to compute the terms resulting from the $(x^2+9)^2$ factor. Thus, introduce the auxiliary fraction \[ \frac{1}{(x^2+t)(x^2+9)^2} \] and expand it as \begin{equation} \label{example3} \frac{1}{(x^2+t)(x^2+9)^2} = \frac{Ax + B}{x^2+t} + \frac{q(x,t)}{(x^2+9)^2}. \end{equation}

We determine $A$ and $B$ following the procedure in The case of a simple complex root. Thus, we clear the identity \eqref{example3} of fractions \[ 1 = (Ax + B) (x^2+9)^2 + q(x,t) (x^2+t) \] then replace $x^2$ by $-t$. We get \[ 1 = (Ax + B) (9-t)^2 \] whence \[ A = 0, \qquad B = \frac{1}{(t-9)^2}, \] and \[ A' = 0, \qquad B' = - \frac{2}{(t-9)^3}. \] At $t=1$ we have: \[ A = 0, \quad A' = 0, \quad B = \frac{1}{64}, \quad B' = \frac{1}{256}. \quad \]

Going back to \eqref{example3}, we differentiate with respect to $t$: \[ -\frac{1}{(x^2+t)^2(x^2+9)^2} = \frac{A'(t)x + B'(t)}{x^2+t} - \frac{A(t)x + B(t)}{(x^2+t)^2} + \frac{q_t(x,t)}{(x^2+9)^2} \] then substitute $t=1$, \[ -\frac{1}{(x^2+1)^2(x^2+9)^2} = \frac{(1/256)}{x^2+1} - \frac{(1/64)}{(x^2+1)^2} + \frac{q_t(x,t)}{(x^2+9)^2}, \] and multiply through by $-1$: \[ \frac{1}{(x^2+1)^2(x^2+9)^2} = - \frac{1}{256(x^2+1)} + \frac{1}{64(x^2+1)^2} - \frac{q_t(x,t)}{(x^2+9)^2}. \]

In a similar way, we may compute the term in the expansion due to the factor $x^2+9$. We introduce the auxiliary fraction \[ \frac{1}{(x^2+1)^2(x^2+t)} \] and expand it as: \begin{equation} \label{example4} \frac{1}{(x^2+1)^2(x^2+t)} = \frac{Ax+B}{x^2+t} + \frac{q(t)}{(x^2+1)^2}. \end{equation}

Upon clearing the identity \eqref{example4} of fractions, we get: \[ 1 = (Ax+B)(x^2+1)^2 + q(t)(x^2+t). \] Then we replace $x^2$ by $-t$: \[ 1 = (Ax+B)(-t+1)^2, \] from which get conclude that \[ A = 0, \quad B = \frac{1}{(t-1)^2}, A' = 0, \quad B' = -\frac{2}{(t-1)^3}, \] whence at $t=9$: \[ A = 0, \quad A' = 0, \quad B = \frac{1}{4}, \quad B' = -\frac{1}{256}. \] Next, we differentiate \eqref{example4} with respect to $t$: \[ -\frac{1}{(x^2+1)^2(x^2+t)^2} = \frac{A'(t)x+B'(t)}{x^2+t} - \frac{A(t)x+B(t)}{(x^2+t)^2} + \frac{q'(t)}{(x^2+1)^2} \] then let $t=9$: \[ -\frac{1}{(x^2+1)^2(x^2+9)^2} = \frac{(-1/256)}{x^2+9} - \frac{(1/64)}{(x^2+9)^2} + \frac{q'(9)}{(x^2+1)^2}, \] and multiply through by $-1$: \[ \frac{1}{(x^2+1)^2(x^2+9)^2} = \frac{1}{256(x^2+9)} + \frac{1}{64(x^2+9)^2} - \frac{q'(9)}{(x^2+1)^2}, \] Putting this together with the earlier result, we conclude that: \[ \frac{1}{(x^2+1)^2(x^2+9)^2} = - \frac{1}{256(x^2+1)} + \frac{1}{64(x^2+1)^2} + \frac{1}{256(x^2+9)} + \frac{1}{64(x^2+9)^2}. \]

An alternative solution to the example above is to let $u=x^2$ and expand the fraction \[ \frac{1}{(u+1)^2(u+9)^2} \] through the method described in The case of repeated real roots.

Test Yourself

  1. Show that \[ \frac{1}{(x^2+1)(x^2+x+1)^2} = - \frac{1}{x^2+1} + \frac{1}{x^2+x+1} + \frac{x+1}{(x^2+x+1)^2} \]
  2. Show that \[ \frac{1}{(x^2+1)^3(x^2+x+1)} = \frac{1}{x^2+1} - \frac{x-1}{(x^2+1)^2} - \frac{x}{(x^2+1)^3} - \frac{1}{x^2+x+1}. \]
  3. Show that \[ \frac{1}{(x+1)^2(x^2+1)^2} = \frac{1}{4(x+1)^2} + \frac{1}{2(x+1)} - \frac{x}{2(x^2+1)^2} - \frac{2x-1}{4(x^2+1)^2}. \] 
 
Author: Rouben Rostamian