# How do you express  x^2/(x^2+9)^2 in partial fractions?

##### 1 Answer
Apr 13, 2016

${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{1}{{x}^{2} + 9} - \frac{9}{{x}^{2} + 9} ^ 2$

#### Explanation:

Partial fractions of ${x}^{2} / {\left({x}^{2} + 9\right)}^{2}$ will be of type

${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{A x + B}{{x}^{2} + 9} + \frac{C x + D}{{x}^{2} + 9} ^ 2$

Simplifying RHS

${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{\left(A x + B\right) \left({x}^{2} + 9\right) + C x + D}{{x}^{2} + 9} ^ 2$ or

${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{A {x}^{3} + B {x}^{2} + 9 A x + 9 B + C x + D}{{x}^{2} + 9} ^ 2$

Now comparing like terms on each side

$A = 0$, $B = 1$, $9 A + C = 0$ and $9 B + D = 0$

As $A = 0$ and $B = 1$, putting these values we get $C = 0$ and $D = - 9$.

Hence ${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{1}{{x}^{2} + 9} - \frac{9}{{x}^{2} + 9} ^ 2$