Question

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

Answer 1

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!