Question

Write the partial fraction decomposition of the rational expression.

$\frac{x^2}{{\left(x^2+4\right)}^3}$

Short answer

The partial fraction decomposition of the given rational expression is given below,

$\frac{x^2}{{\left(x^2+4\right)}^3}=\frac{1}{{\left(x^2+4\right)}^2}-\frac{4}{{\left(x^2+4\right)}^3}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\frac{x^2}{{\left(x^2+4\right)}^3}$

Step 2. Decompose the rational expression.

The denominator has a repeated non-reducible quadratic factor $\left(x^2+4\right)$. Then we decompose the rational expression as follows,

$\frac{x^2}{{\left(x^2+4\right)}^3}=\frac{Ax+B}{\left(x^2+4\right)}+\frac{Cx+D}{{\left(x^2+4\right)}^2}+\frac{Ex+F}{{\left(x^2+4\right)}^3}......\left(i\right)$

Where, A,B,C,D,E and F are to be determined.

Multiply every side by ${\left(x^2+4\right)}^3$,

$x^2=\left(Ax+B\right){\left(x^2+4\right)}^2+\left(Cx+D\right)\left(x^2+4\right)+\left(Ex+F\right)......\left(ii\right)$

Step 3. Expand the identity.

On expanding the identity in equation (ii),

$x^2=x^5\left(A\right)+x^4\left(B\right)+x^3\left(C\right)+x^2\left(8A-8B+D\right)+x\left(16A+4C+E\right)+16B+4D+F$

Equate the coefficients of like powers of x to get,

$$\begin{gathered} A=0 \\ B=0 \\ C=0 \\ 8A+8B+D=1 \\ 10A+4C+E=0 \\ 16B+4D+F=0 \end{gathered}$$

Step 4. Substitute the values in equation (i).

On solving the system of equations, we get,

$$\begin{gathered} A=0, \\ B=0, \\ C=0, \\ D=1, \\ E=0, \\ F=-4 \end{gathered}$$

Substitute the values in equation (i),

$\frac{x^2}{{\left(x^2+4\right)}^3}=\frac{1}{{\left(x^2+4\right)}^2}-\frac{4}{{\left(x^2+4\right)}^3}$