# How do you integrate  (3x^2 + 12x - 20 )/(x^4 - 8x^2 + 16) using partial fractions?

Dec 7, 2016

The answer is =2/(x+2)-ln(∣x+2∣)-1/(x-2)+ln(∣x-2∣)+C

#### Explanation:

We use,

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

and ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

to factorise the denominator

${x}^{4} - 8 {x}^{2} + 16 = {\left({x}^{2} - 4\right)}^{2}$

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

So,

$\frac{3 {x}^{2} + 12 x - 20}{{x}^{4} - 8 {x}^{2} + 16} = \frac{3 {x}^{2} + 12 x - 20}{{\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}}$

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

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

So,

$3 {x}^{2} + 12 x - 20 = A {\left(x - 2\right)}^{2} + B \left(x + 2\right) {\left(x - 2\right)}^{2} + C {\left(x + 2\right)}^{2} + D \left(x - 2\right) {\left(x + 2\right)}^{2}$

Let $x = 2$, $\implies$, $16 = 16 C$, $\implies$, $C = 1$

Let $x = - 2$, $\implies$, $- 32 = 16 A$, $\implies$, $A = - 2$

Coefficients of ${x}^{3}$, $0 = B + D$

And $- 20 = 4 A + 8 B + 4 C - 8 D$

$- 20 = - 8 + 8 B + 4 - 8 D$

$- 16 = 8 B - 8 D$

$B - D = - 2$

So, $B = - 1$ and $D = 1$

$\int \frac{\left(5 {x}^{3} + 2 {x}^{2} - 12 x - 8\right) \mathrm{dx}}{{x}^{4} - 8 {x}^{2} + 16}$

$= = \int \frac{- 2 \mathrm{dx}}{x + 2} ^ 2 + \int \frac{- 1 \mathrm{dx}}{x + 2} + \int \frac{1 \mathrm{dx}}{x - 2} ^ 2 + \int \frac{1 \mathrm{dx}}{x - 2}$

=2/(x+2)-ln(∣x+2∣)-1/(x-2)+ln(∣x-2∣)+C