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

1 Answer

int (4x^2+6x-2)/((x-1)(x+1)^2) dx=
2ln(x-1 )+2ln(x+1)-2/(x+1)+C_o

Explanation:

Set up the equation to solve for the variables A, B,C
int (4x^2+6x-2)/((x-1)(x+1)^2) dx=int (A/(x-1 )+B/(x+1)+C/(x+1)^2)dx

Let us solve for A, B, C first

(4x^2+6x-2)/((x-1)(x+1)^2) =A/(x-1 )+B/(x+1)+C/(x+1)^2

LCD =(x-1)(x+1)^2

(4x^2+6x-2)/((x-1)(x+1)^2) =(A(x+1)^2+B(x^2-1)+C(x-1))/((x-1)(x+1)^2)

Simplify

(4x^2+6x-2)/((x-1)(x+1)^2) =(A(x^2+2x+1)+B(x^2-1)+C(x-1))/((x-1)(x+1)^2)

(4x^2+6x-2)/((x-1)(x+1)^2) =(Ax^2+2Ax+A+Bx^2-B+Cx-C)/((x-1)(x+1)^2)

Rearrange the terms of the right side

(4x^2+6x-2)/((x-1)(x+1)^2) =(Ax^2+Bx^2+2Ax+Cx+A-B-C)/((x-1)(x+1)^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" "first equation
2A+C=6" "second equation
A-B-C=-2" "third equation

Simultaneous solution using second and third equation results to

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

3A-B=4" "fourth equation

Using now the first and the fourth equations

3A-B=4" "fourth equation
3(4-B)-B=4" "fourth equation

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

Solve for A using 3A-B=4" "fourth equation
3A-2=4" "fourth equation
3A=4+2
3A=6
A=2

Solve C using the 2A+C=6" "second equation and A=2 and B=2

2A+C=6" "second equation
2(2)+C=6
4+C=6
C=6-4
C=2

We now perform our integration
int (4x^2+6x-2)/((x-1)(x+1)^2) dx=int (2/(x-1 )+2/(x+1)+2/(x+1)^2)dx

int (4x^2+6x-2)/((x-1)(x+1)^2) dx=int (2/(x-1 )+2/(x+1)+2*(x+1)^(-2))dx

int (4x^2+6x-2)/((x-1)(x+1)^2) dx=2ln(x-1 )+2ln(x+1)+(2*(x+1)^(-2+1))/(-2+1)+C_o

int (4x^2+6x-2)/((x-1)(x+1)^2) dx=2ln(x-1 )+2ln(x+1)-2/(x+1)+C_o

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