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

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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!