How do you integrate $(3x+11)/(x^2-x-6)$ using partial fractions?

Question

8066 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-19T15:43:30

The answer is $4ln|x- 3| - ln|x+ 2| + C$

The denominator can be factored as $(x - 3)(x + 2)$ , so:

$A/(x- 3) + B/(x + 2) = (3x + 11)/((x- 3)(x + 2))$

$A(x + 2) + B(x- 3) = 3x +11$

$Ax + 2A + Bx - 3B = 3x +11$

$(A + B)x + (2A - 3B) = 3x + 11$

We now write a systems of equations to solve for $A$ and $B$ .

${(A + B = 3), (2A - 3B = 11):}$

Solving:

$B = 3 - A -> 2A - 3(3 - A) = 11 -> 2A - 9 + 3A = 11 ->5A = 20-> A= 4$

$A + B = 3 -> 4 + B = 3 -> B = -1$

So, $A = 4$ and $B = -1$ .

The partial fraction decomposition is therefore $color(blue)(4/(x- 3) - 1/(x + 2))$ .

This can be integrated using the rule $int1/xdx = ln|x| + C$ .

$int(4/(x- 3) - 1/(x + 2)) = 4ln|x- 3| - ln|x+ 2| + C$

Hopefully this helps!

How do you integrate (3x+11)/(x^2-x-6)(3x+11)/(x^2-x-6) using partial fractions?