# How do you integrate int (5x^3+2x^2-12x-8)/(x^4-8x^2+16) dx using partial fractions?

Dec 4, 2016

The answer is =1/(x+2)+2ln(∣x+2∣)-1/(x-2)+3ln(∣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,

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

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

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

So,

$5 {x}^{3} + 2 {x}^{2} - 12 x - 8 = A {\left(x - 2\right)}^{2} + B \left(x + 2\right) {\left(x - 2\right)}^{2} + C {\left(x + 2\right)}^{2} + D \left(x - 2\right) {\left(x + 2\right)}^{2}$

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

Let $x = - 2$, $\implies$, $- 16 = 16 A$, $\implies$, $A = - 1$

Coefficients of ${x}^{3}$, $5 = B + D$

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

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

$- 8 = 8 B - 8 D$

$B - D = - 1$

So, $B = 2$ and $D = 3$

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

$= = \int \frac{- 1 \mathrm{dx}}{x + 2} ^ 2 + \int \frac{2 \mathrm{dx}}{x + 2} + \int \frac{1 \mathrm{dx}}{x - 2} ^ 2 + \int \frac{3 \mathrm{dx}}{x - 2}$

=1/(x+2)+2ln(∣x+2∣)-1/(x-2)+3ln(∣x-2∣)+C