How do you integrate #int (3x^2 + 10x -5)/ ( (x+1)^2(x-2) )# using partial fractions?

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Answer 1 · 2017-07-19T10:56:05

#int(3x^2+10x-5)/((x+1)^2(x-2))dx=-4/(x+1)+3ln|x-2| +C#

Let's start with :

#(3x^2+10x-5)/((x+1)^2(x-2))=A/(x+1)+B/(x+1)^2+C/(x-2)=#

#(A(x+1)(x-2)+B(x-2)+C(x+1)^2)/((x+1)^2(x-2))=#

#(A(x^2-x-2)+Bx-2B+C(x^2+2x+1))/((x+1)^2(x-2))=#

#(Ax^2-Ax-2A+Bx-2B+Cx^2+2Cx+C)/((x+1)^2(x-2))=#

#((A+C)x^2+(-A+B+2C)x+(-2A-2B+C))/((x+1)^2(x-2))#

So we know now :

#A+C=3#
#-A+B+2C=10#
#-2A-2B+C=-5#

From this system we get :

#A=0, B=4, C=3#

Let's get to the integral now:

#int(3x^2+10x-5)/((x+1)^2(x-2))dx=int(4/(x+1)^2+3/(x-2))dx=#

#int4/(x+1)^2dx+int3/(x-2)dx=#

#int4(x+1)^(-2)dx+3ln|x-2|=#

#-4(x+1)^(-1)+3ln|x-2|=#

#-4/(x+1)+3ln|x-2| +C#