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

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

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Answer 1 · 2016-03-29T23:50:14

$\int \frac{4 x^{2} + 6 x - 2}{(x - 1) {(x + 1)}^{2}} d x =$ $int (4x^2+6x-2)/((x-1)(x+1)^2) dx=$
$2 ln (x - 1) + 2 ln (x + 1) - \frac{2}{x + 1} + C_{o}$ $2ln(x-1 )+2ln(x+1)-2/(x+1)+C_o$

Explanation:

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

Let us solve for A, B, C first

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

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

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

Simplify

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

$\frac{4 x^{2} + 6 x - 2}{(x - 1) {(x + 1)}^{2}} = \frac{A x^{2} + 2 A x + A + B x^{2} - B + C x - C}{(x - 1) {(x + 1)}^{2}}$ $(4x^2+6x-2)/((x-1)(x+1)^2) =(Ax^2+2Ax+A+Bx^2-B+Cx-C)/((x-1)(x+1)^2)$

Rearrange the terms of the right side

$\frac{4 x^{2} + 6 x - 2}{(x - 1) {(x + 1)}^{2}} = \frac{A x^{2} + B x^{2} + 2 A x + C x + A - B - C}{(x - 1) {(x + 1)}^{2}}$ $(4x^2+6x-2)/((x-1)(x+1)^2) =(Ax^2+Bx^2+2Ax+Cx+A-B-C)/((x-1)(x+1)^2)$

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

$A + B = 4$ $A+B=4" "$ first equation
$2 A + C = 6$ $2A+C=6" "$ second equation
$A - B - C = - 2$ $A-B-C=-2" "$ third equation

Simultaneous solution using second and third equation results to

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

$3 A - B = 4$ $3A-B=4" "$ fourth equation

Using now the first and the fourth equations

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

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

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

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

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

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

$\int \frac{4 x^{2} + 6 x - 2}{(x - 1) {(x + 1)}^{2}} d x = \int (\frac{2}{x - 1} + \frac{2}{x + 1} + 2 \cdot {(x + 1)}^{- 2}) d x$ $int (4x^2+6x-2)/((x-1)(x+1)^2) dx=int (2/(x-1 )+2/(x+1)+2*(x+1)^(-2))dx$

$\int \frac{4 x^{2} + 6 x - 2}{(x - 1) {(x + 1)}^{2}} d x = 2 ln (x - 1) + 2 ln (x + 1) + \frac{2 \cdot {(x + 1)}^{- 2 + 1}}{- 2 + 1} + C_{o}$ $int (4x^2+6x-2)/((x-1)(x+1)^2) dx=2ln(x-1 )+2ln(x+1)+(2*(x+1)^(-2+1))/(-2+1)+C_o$

$\int \frac{4 x^{2} + 6 x - 2}{(x - 1) {(x + 1)}^{2}} d x = 2 ln (x - 1) + 2 ln (x + 1) - \frac{2}{x + 1} + C_{o}$ $int (4x^2+6x-2)/((x-1)(x+1)^2) dx=2ln(x-1 )+2ln(x+1)-2/(x+1)+C_o$

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

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

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