Integration by partial fractions?

First divide to get a quotient and a remainder where the remainder has degree less than that of the denominator. In this case, you will be dividing 3x^2-6 by x^2-x-2 to get 3+3x/(x+1)(x-2). Now do the partial fractions decomposition on the second piece. So you want to write

3x/(x+1)(x-2)=A/(x+1) +B/(x-2).

Multiply by (x+1)(x-2) to get

3x=A(x-2)+B(x+1).

Let x=2 to find B=2 and let x=-1 to find A=1.

Thus, you are integrating

3+ 1/(x+1) +2/(x-2).

Let (3x^2-6)/[(x+1)(x-2)] = A/(x+1) + B/(x-2), where A and B are constants to be determined.

Multiply both sides by the original denominator, and get

3x^2-6 = A(x-2) + B(x+1).

Now, use the "method of creating zeros". Let x=2. Then the term with A will drop out, and you can solve for B. Then, go back to the equation above, plug in x=-1, and the term with B will drop out, and you can get A.

Once you have A and B, you can integrate A/(x+1) + B/(x-2), using the rule

integral of 1/(x+a) dx = ln |x+a|

and you're done.

integral of 3x^2-6/(x+1)(x-2)=3x^2-6/x^2-x-2

dividing x^2-x-2)3x^2 -6(3

3x^2-3x-6

----------------

3x

so the expression reduces to 3+[3x/(x+1)(x-2)]

let A/(x+1)+B/(x-2)=3x/(x+1)(x-2)

A(x-2)+B(x+1)=3x

comparing the coefficients

A+B=3

-2A+B=0

3A=3 so A=1

substituting B=2

now the integral is [3+{1/(x+1)}+{2/(x-2)}]dx

=3x+ln(x+1)(x-2)^2+C