How do you integrate #(2x^3-3x^2+x+1)/(-2x+1)#?

1 Answer
Sasha P.
Sep 14, 2015

#-x^3/3+x^2/2-1/2lnabs(2x-1)+C#

Explanation:

We have to divide polinomials:

#(2x^3-3x^2+x+1):(-2x+1)=-x^2+x#
#-2x^3+x^2#
#----#
#-2x^2+x+1#
#+2x^2-x#
#----#
#1#

#(2x^3-3x^2+x+1)/(-2x+1)=-x^2+x+1/(-2x+1)#
#int(2x^3-3x^2+x+1)/(-2x+1)dx=#
#=int(-x^2+x+1/(-2x+1))dx=#
#=-intx^2dx+intxdx-intdx/(2x-1)=I#
#2x-1=t, 2dx=dt, dx=dt/2#
#I=-x^3/3+x^2/2-intdt/2 1/t= #
#=-x^3/3+x^2/2-1/2lnabs(2x-1)+C#

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