First divide the numerator by the denominator using long division.
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So, our new integral is int(2x + (x + 5)/(x^2 - 2x - 8))dx
We now can write (x + 5)/(x^2 - 2x - 8) in partial fractions.
First of all, x^2 - 2x - 8 can be factored as (x- 4)(x + 2).
A/(x- 4) + B/(x + 2) = (x + 5)/((x- 4)(x + 2))
A(x + 2) + B(x- 4) = x + 5
Ax + 2A + Bx - 4B = x + 5
(A + B)x + (2A - 4B) = x + 5
We can now write a system of equations.
{(A + B = 1), (2A - 4B = 5):}
Solving:
B = 1 - A
2A - 4(1 - A) = 5
2A - 4 + 4A = 5
6A = 9
A = 3/2
B = -1/2
Hence, the partial fraction decomposition of (x + 5)/(x^2- 2x - 8) is 3/(2(x - 4)) - 1/(2(x + 2))
The integral now becomes:
int(2x + 3/(2(x - 4)) - 1/(2(x + 2)))dx
Which can be readily integrated as;
x^2 + 3/2ln|x - 4| - 1/2ln|x + 2| + C using the rule int(1/u)du = ln|u| + C and int(x^n)dx = (x^(n + 1))/(n + 1). You can check the result by differentiating.
Hopefully this helps!
