How do you integrate #int (3x^3+2x^2-7x-6)/(x^2-4) dx# using partial fractions?

1 Answer
Nov 10, 2016

The integral is #y =3/2x^2 + 2x + 2ln|x + 2| + 3ln|x - 2| + C#

Explanation:

Start by using long division to divide. We want the degree of the numerator to be less than the degree of the denominator.

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So, the quotient is #3x + 2# with a remainder of #5x + 2#, which can be written as #3x + 2 + (5x + 2)/(x^2 -4)#.

We can now write #(5x +2)/(x^2 - 4)# in partial fractions.

#A/(x + 2) + B/(x - 2) = (5x + 2)/((x + 2)(x- 2))#

#(A(x- 2)) + B(x + 2)) = 5x + 2#

#Ax - 2A + Bx + 2B = 5x + 2#

#(A + B)x + (2B - 2A) = 5x + 2#

So, #A+ B = 5# and #2B - 2A = 2#.

#B = 5 - A#

#2(5 - A) - 2A = 2#

#10 - 2A - 2A = 2#

#-4A = -8#

#A = 2#

#:.2 + B = 5#

#B = 3#

So, the partial fraction decomposition is #2/(x + 2) + 3/(x - 2)#.

We rewrite the integral like this:

#=int(3x + 2 + 2/(x + 2) + 3/(x- 2))dx#

Integrate using the rule #int(1/x)dx = lnx + C#.

#=3/2x^2 + 2x + 2ln|x + 2| + 3ln|x - 2| + C#

You can check the answer by differentiating. You will get the initial problem.

Hopefully this helps!