# How do you express (x+1)/[(x^2+1)^2(x^2)] in partial fractions?

Jan 20, 2017

The answer is $= \frac{1}{x} ^ 2 + \frac{1}{x} + \frac{- x - 1}{{x}^{2} + 1} ^ 2 + \frac{- x - 1}{{x}^{2} + 1}$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{x + 1}{{\left({x}^{2} + 1\right)}^{2} {x}^{2}} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C x + D}{{x}^{2} + 1} ^ 2 + \frac{E x + F}{{x}^{2} + 1}$

$= \frac{A {\left({x}^{2} + 1\right)}^{2} + B \left(x\right) {\left({x}^{2} + 1\right)}^{2} + \left(C x + D\right) {x}^{2} + \left(E x + F\right) \left({x}^{2} + 1\right) {x}^{2}}{{\left({x}^{2} + 1\right)}^{2} {x}^{2}}$

Equalising the numerators

$x + 1 = A {\left({x}^{2} + 1\right)}^{2} + B \left(x\right) {\left({x}^{2} + 1\right)}^{2} + \left(C x + D\right) {x}^{2} + \left(E x + F\right) \left({x}^{2} + 1\right) {x}^{2}$

Let $x = 0$, $\implies$, $1 = A$

Coefficients of $x$, $1 = B$

Coefficients of ${x}^{2}$, $0 = 2 A + D + F$

Coefficients of ${x}^{3}$, $0 = 2 B + C + E$

coefficients of ${x}^{4}$, $0 = A + F$, $\implies$, $F = - A = - 1$

$D = - 2 A - F = - 2 + 1 = - 1$

Coefficients of ${x}^{5}$, $0 = B + E$, $\implies$, $E = - B = - 1$

$C = - 2 B - E = - 2 + 1 = - 1$

Therefore,

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