How do you use partial fraction decomposition to decompose the fraction to integrate #11/(3x^2-7x-6)#?

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Answer 1 · 2018-04-04T08:48:20

#int 11/(3x^2-7x-6)dx = ln abs((x-3)/(3x+2))+C#

Factorize the denominator by solving the equation:

#3x^2-7x-6 = 0#

#x = (7+-sqrt(49+72))/6 = (7+-sqrt(121))/6 = (7+-11)/6={(-2/3),(3):}#

Then:

#3x^2-7x-6 = 3(x+2/3)(x-3) = (3x+2)(x-3)#

Apply partial fractions decomposition:

#11/(3x^2-7x-6) = A/(3x+2)+B/(x-3)#

#11/(3x^2-7x-6) = (A(x-3)+B(3x+2))/(3x^2-7x-6)#

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

#{(A+3B = 0),(-3A+2B=11):}#

#{(A=-3),(B=1):}#

So:

#11/(3x^2-7x-6) = -3/(3x+2)+1/(x-3)#

#int 11/(3x^2-7x-6)dx = -int (3dx)/(3x+2)+int dx/(x-3)#

#int 11/(3x^2-7x-6)dx = -int (d(3x+2))/(3x+2)+int (d(x-3))/(x-3)#

#int 11/(3x^2-7x-6)dx = -ln abs(3x+2)+ln abs(x-3)+C#

#int 11/(3x^2-7x-6)dx = ln abs((x-3)/(3x+2))+C#