Answer
$\dfrac{x^2+2x+3}{(x^2+4)^2}=\dfrac{x}{x^2+4}+\dfrac{2x-1}{(x^2+4)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{x^2+2x+3}{(x^2+4)^2}$
As the denominator is factored, we can write the partial fraction decomposition:
$\dfrac{x^2+2x+3}{(x^2+4)^2}=\dfrac{Ax+B}{x^2+4}+\dfrac{Cx+D}{(x^2+4)^2}$
Multiply all terms by the least common denominator $(x^2+4)^2$:
$(x^2+4)^2\cdot\dfrac{x^2+2x+3}{(x^2+4)^2}=(x^2+4)^2\cdot\dfrac{Ax+B}{x^2+4}+(x^2+4)^2\cdot\dfrac{Cx+D}{(x^2+4)^2}$
$x^2+2x+3=(Ax+B)(x^2+4)+(Cx+D)$
$x^2+2x+3=Ax^3+4Ax+Bx^2+4B+Cx+D$
$x^2+2x+3=Ax^3+Bx^2+(4A+C)x+(4B+D)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A=0\\
B=1\\
4A+C=2\\
4B+D=3
\end{cases}$
Solve the system:
$A=0$
$B=1$
$4A+C=2$
$4(0)+C=2$
$C=2$
$4B+D=3$
$4(1)+D=3$
$4+D=3$
$D=-1$
The partial fraction decomposition is:
$\dfrac{x^2+2x+3}{(x^2+4)^2}=\dfrac{x}{x^2+4}+\dfrac{2x-1}{(x^2+4)^2}$
