Partial Fractions with a Repeated and a Irreducible Quadratic factor

Solution 1:

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 $\,\color{#c0d}{w^2\!+\!1 = -2w}$

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

Solution 2:

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 $D=1$