# How do you integrate 5/(x^2-1)^2 using partial fractions?

Nov 27, 2017

$\int \frac{5}{{x}^{2} - 1} ^ 2 \cdot \mathrm{dx} = \frac{5}{4} \cdot L n \left(\frac{x + 1}{x - 1}\right) - \frac{5}{2} \cdot \frac{x}{{x}^{2} - 1} + C$

#### Explanation:

$\int \frac{5}{{x}^{2} - 1} ^ 2 \cdot \mathrm{dx}$

=$\int \frac{5}{\left(x + 1\right) \cdot \left(x - 1\right)} ^ 2 \cdot \mathrm{dx}$

I decomposed integrand into basic fractions,

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

After expanding denominator,

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

Set $x = - 1$, so $4 B = 5$ or $B = \frac{5}{4}$

Set $x = 1$, so $4 D = 5$ or $D = \frac{5}{4}$

Due to denominator have double roots of $- 1$ and $1$, I took derivative both sides,

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

Set $x = - 1$, so $4 A - 4 B = 0$ or $A = B = \frac{5}{4}$

Set $x = 1$, so $4 C + 4 D = 0$ or $C = - D = - \frac{5}{4}$

Hence,

$\int \frac{5}{\left(x + 1\right) \cdot \left(x - 1\right)} ^ 2 \cdot \mathrm{dx} = \frac{5}{4} \cdot \int \frac{\mathrm{dx}}{x + 1} + \frac{5}{4} \cdot \int \frac{\mathrm{dx}}{x + 1} ^ 2 - \frac{5}{4} \cdot \int \frac{\mathrm{dx}}{x - 1} + \frac{5}{4} \cdot \int \frac{\mathrm{dx}}{x - 1} ^ 2$

=$\frac{5}{4} \cdot L n \left(x + 1\right) - \frac{5}{4} \cdot {\left(x + 1\right)}^{- 1} - \frac{5}{4} \cdot L n \left(x - 1\right) - \frac{5}{4} \cdot {\left(x - 1\right)}^{- 1} + C$

=$\frac{5}{4} \cdot L n \left(\frac{x + 1}{x - 1}\right) - \frac{5}{2} \cdot \frac{x}{{x}^{2} - 1} + C$