# How do you express (2x^2+5)/(x^2+1)^2 in partial fractions?

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} ^ 2 = \frac{3}{{x}^{2} + 1} ^ 2 + \frac{2}{{x}^{2} + 1}$

#### Explanation:

Begin with the set up of the variables A, B, C, D

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} ^ 2 = \frac{A x + B}{{x}^{2} + 1} ^ 2 + \frac{C x + D}{{x}^{2} + 1}$

the LCD(Least common Denominator) is $= {\left({x}^{2} + 1\right)}^{2}$

so that the right side of the equation becomes

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} ^ 2 = \frac{A x + B}{{x}^{2} + 1} ^ 2 + \frac{\left(C x + D\right) \left({x}^{2} + 1\right)}{{x}^{2} + 1} ^ 2$

expand

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} ^ 2 = \frac{A x + B}{{x}^{2} + 1} ^ 2 + \frac{C {x}^{3} + C x + D {x}^{2} + D}{{x}^{2} + 1} ^ 2$

combine

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} ^ 2 = \frac{A x + B + C {x}^{3} + C x + D {x}^{2} + D}{{x}^{2} + 1} ^ 2$

rearrange

$\frac{0 \cdot {x}^{3} + 2 {x}^{2} + 0 \cdot x + 5 \cdot {x}^{0}}{{x}^{2} + 1} ^ 2 = \frac{C {x}^{3} + D {x}^{2} + \left(A + C\right) x + \left(B + D\right) {x}^{0}}{{x}^{2} + 1} ^ 2$

the equations can now be found by equating the numerical coefficients:

$C = 0$
$D = 2$
$A + C = 0$
$B + D = 5$

Solving for A, B, C, D:

$A = 0$ and $B = 3$ and $C = 0$ and $D = 2$

so that the partial fractions are

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} ^ 2 = \frac{A x + B}{{x}^{2} + 1} ^ 2 + \frac{C x + D}{{x}^{2} + 1} = \frac{0 \cdot x + 3}{{x}^{2} + 1} ^ 2 + \frac{0 \cdot x + 2}{{x}^{2} + 1}$

and

$\frac{2 {x}^{2} + 5}{{x}^{2} + 1} ^ 2 = \frac{3}{{x}^{2} + 1} ^ 2 + \frac{2}{{x}^{2} + 1}$

have a nice day! from the Philippines..