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

Question

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

Charles Arocha · Accepted Answer

#int (4x^2+6x-2)/((x-1)(x+1)^2) dx=#
#2ln(x-1 )+2ln(x+1)-2/(x+1)+C_o# 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# First, let's solve for A, B, and C. #(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 right side's terms. #(4x^2+6x-2)/((x-1)(x+1)^2) =(Ax^2+Bx^2+2Ax+Cx+A-B-C)/((x-1)(x+1)^2)# Let's align the left and right terms' numerical coefficients to set up the equations for A, B, and C to solve. #A+B=4" "#first equation #2A+C=6" "#second equation #A-B-C=-2" "#third equation Using the second and third equations simultaneously leads to #2A+A+C-C-B=6-2# #3A-B=4" "#fourth equation Applying the first and fourth equations at this time #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# May God bless you all. I hope this explanation helps.

Charles Arocha · Answer

undefined To integrate the given rational function using partial fractions, you first need to decompose it into partial fractions. Start by factoring the denominator completely:

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

The partial fraction decomposition will have the following form:

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

Next, find the values of A, B, and C by equating coefficients of like terms in the original expression and the decomposition.

Multiply both sides of the equation by the denominator of the original expression to eliminate fractions:

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

Now, substitute suitable values of x to solve for A, B, and C. Typically, you would choose values that make some terms drop out or simplify. A common approach is to choose values such as x = 1, x = -1, and x = 0.

After finding the values of A, B, and C, rewrite the original expression as the sum of the partial fractions. Then integrate each term separately.

Finally, integrate each term, and you will get the result.