How do you integrate #(3x^3-5x^2-11x+9)/(x^2-2x-3)# using partial fractions?

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Answer 1 · 2017-07-31T14:14:00

#int(3x^3-5x^2-11x+9)/(x^2-2x-3)dx#

= #(3x^2)/2+x+3ln|x-3|-3ln|x+1|+c#

For converting to partial fractions, we must have degreeof numerator less than that of denominator. Hence as

#3x^3-5x^2-11x+9=3x(x^2-2x-3)+1(x^2-2x-3)+12# and hence

#(3x^3-5x^2-11x+9)/(x^2-2x-3)=3x+1+12/(x^2-2x-3)#

As #(x^2-2x-3)=(x-3)(x+1)#, let

#12/(x^2-2x-3)hArrA/(x-3)+B/(x+1)=(A(x+1)+B(x-3))/(x^2-2x-3)#

if we put #x=3#, #4A=12# i.e. #A=3#

and if we put #x=-1#, #-4B=12# or #B=-3#

Hence #(3x^3-5x^2-11x+9)/(x^2-2x-3)=3x+1+12/(x^2-2x-3)#

and #int(3x^3-5x^2-11x+9)/(x^2-2x-3)dx#

= #int(3x+1+3/(x-3)-3/(x+1))dx#

= #(3x^2)/2+x+3ln|x-3|-3ln|x+1|+c#