# Partial Fractions with a Repeated and a Irreducible Quadratic factor

I am trying to make this into a partial fractions form but i can't seem to find a way to do it.

The question is here:

Change into a partial fractions form.

\begin{align} \frac{2s}{(s+1)^2(s^2 + 1)} \\\\ \end{align}

i change to \begin{align} \frac{A}{(s+1)}+\frac{B}{(s+1)^2}+\frac{Cs+D}{(s^2+1)} \end{align}

I expand it and group like terms tgt but i am stump at this point \begin{align} 2s = (A+C)s^3 + (B+A+2C+D)s^2 + (A+C+2D)s + (B+A+D) \end{align}

Anyone can help?

## 4 Answers

Since you arrived to \begin{align} 2s = (A+C)s^3 + (B+A+2C+D)s^2 + (A+C+2D)s + (B+A+D) \end{align} identification of the coefficients of the different powers of $s$ gives $$A+C=0$$ $$B+A+2C+D=0$$ $$A+C+2D=2$$ $$B+A+D=0$$ for which the solutions are easily obtained either using matrix or elimination and the final result is : $A=0$,$B=-1$,$C=0$ and $D=1$

• You are very welcome ! – Claude Leibovici May 21 '14 at 9:43
• @user152399 It's much easier to use Heaviside - see my answer. – Bill Dubuque May 21 '14 at 15:09
• @BillDubuque. You are right but, by the end, Jyrki Lahtonen's solution is the fastest. Cheers. – Claude Leibovici May 22 '14 at 3:46
• @Claude I meant easiest in general (ad-hoc methods don't generally apply). – Bill Dubuque May 22 '14 at 4:39

Such problems are easy using the Heaviside cover up method,. As I show in that answer, the method generalizes to quadratic denominators. Let's apply it to your problem.

$$\dfrac{a}{x+1} + \dfrac{b}{(x+1)^2} + \dfrac{cx+d}{x^2+1}\, =\, \dfrac{2x}{(x+1)^2(x^2+1)}\tag{E}$$

Multiplying $\rm\,(E)\,$ by $\,(x+1)^2,\,$ then evaluating at its root $\,x = -1\,$ yields

$$\ b\, =\, \dfrac{2x}{x^2+1}\Bigg|_{\large x\,=\,-1} =\, -1$$

Subtracting out this known summand leaves a fraction in partial form, so we are done:

$$\ \dfrac{2x}{(x+1)^2(x^2+1)}\, -\, \dfrac{-1}{(x+1)^2}\, =\, \dfrac{2x+x^2+1}{(x+1)^2(x^2+1)}\, =\, \dfrac{1}{x^2+1}\quad {\bf QED}$$

Remark $\$ It is quite instructive to use the quadratic Heaviside method for all terms:

Multiplying $\rm\,(E)\,$ by $\,x^2+1\,$ then evaluating at its root $\,x = i\,$ yields

$$\color{#0a0}c\, i + \color{#c00}d\, =\, \dfrac{2i}{(i\!+\!1)^2} \, =\, \dfrac{2i}{2i} \,=\, \color{#c00}1\,\Rightarrow\ \color{#0a0}{c = 0},\ \color{#c00}{d = 1}$$

Multiplying $\rm\,(E)\,$ by $\,(x+1)^2\,$ then evaluating at its root $\,x = w,\,$ so $\,\color{#c0d}{w^2\!+\!1 = -2w}$

$$a(w+1)+ b\, =\, \dfrac{2w}{\color{#c0d}{w^2\!+\!1}}\,=\,\dfrac{\ \ \,2w}{\color{#c0d}{-2w}} \,=\, -1\,\Rightarrow\ a=0\,\Rightarrow\, b=-1$$

Hint: It is probably easiest to observe that $$2s=(s+1)^2-(s^2+1).$$ Use that in the numerator and see what you can cancel from the two terms you get.

• Sorry, I just love spotting ad hoc tricks. – Jyrki Lahtonen May 21 '14 at 9:31

HINT:

Using Partial Fraction Decomposition formula, $$\frac{2s}{(s+1)^2(s^2+1)}=\frac A{s+1}+\frac B{(s+1)^2}+\frac{Cs+D}{s^2+1}$$

• I tried it but i only am able to find B with Cover up rule. – Alex May 21 '14 at 8:38
• @user152399, If you have four unknowns with four linear equations, right? – lab bhattacharjee May 21 '14 at 9:18
• I don't understand what you mean. – Alex May 21 '14 at 9:23
• @lab Using quadratic Heaviside is usually simpler than solving such systems of equations, see the Remark in my answer. – Bill Dubuque May 21 '14 at 15:44