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

1 Answer
Noah G
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!

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