# How do you integrate int (4x^2+6x-2)/((x-1)(x+1)^2) using partial fractions?

$\int \frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx} =$
$2 \ln \left(x - 1\right) + 2 \ln \left(x + 1\right) - \frac{2}{x + 1} + {C}_{o}$

#### Explanation:

Set up the equation to solve for the variables A, B,C
$\int \frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx} = \int \left(\frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 1} ^ 2\right) \mathrm{dx}$

Let us solve for A, B, C first

$\frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 1} ^ 2$

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

$\frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} = \frac{A {\left(x + 1\right)}^{2} + B \left({x}^{2} - 1\right) + C \left(x - 1\right)}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$

Simplify

$\frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} = \frac{A \left({x}^{2} + 2 x + 1\right) + B \left({x}^{2} - 1\right) + C \left(x - 1\right)}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$

$\frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} = \frac{A {x}^{2} + 2 A x + A + B {x}^{2} - B + C x - C}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$

Rearrange the terms of the right side

$\frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} = \frac{A {x}^{2} + B {x}^{2} + 2 A x + C x + A - B - C}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$

let us set up the equations to solve for A, B, C by matching the numerical coefficients of left and right terms

$A + B = 4 \text{ }$first equation
$2 A + C = 6 \text{ }$second equation
$A - B - C = - 2 \text{ }$third equation

Simultaneous solution using second and third equation results to

$2 A + A + C - C - B = 6 - 2$

$3 A - B = 4 \text{ }$fourth equation

Using now the first and the fourth equations

$3 A - B = 4 \text{ }$fourth equation
$3 \left(4 - B\right) - B = 4 \text{ }$fourth equation

$12 - 3 B - B = 4$
$- 4 B = 4 - 12$
$- 4 B = - 8$
$B = 2$

Solve for A using $3 A - B = 4 \text{ }$fourth equation
$3 A - 2 = 4 \text{ }$fourth equation
$3 A = 4 + 2$
$3 A = 6$
$A = 2$

Solve C using the $2 A + C = 6 \text{ }$second equation and $A = 2$ and $B = 2$

$2 A + C = 6 \text{ }$second equation
$2 \left(2\right) + C = 6$
$4 + C = 6$
$C = 6 - 4$
$C = 2$

We now perform our integration
$\int \frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx} = \int \left(\frac{2}{x - 1} + \frac{2}{x + 1} + \frac{2}{x + 1} ^ 2\right) \mathrm{dx}$

$\int \frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx} = \int \left(\frac{2}{x - 1} + \frac{2}{x + 1} + 2 \cdot {\left(x + 1\right)}^{- 2}\right) \mathrm{dx}$

$\int \frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx} = 2 \ln \left(x - 1\right) + 2 \ln \left(x + 1\right) + \frac{2 \cdot {\left(x + 1\right)}^{- 2 + 1}}{- 2 + 1} + {C}_{o}$

$\int \frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx} = 2 \ln \left(x - 1\right) + 2 \ln \left(x + 1\right) - \frac{2}{x + 1} + {C}_{o}$

God bless.....I hope the explanation is useful.