How do you find the indefinite integral of #int (5x^2 – 4x + 7)/((x^2+1)(2x – 1)) dx#?

1 Answer
Oct 18, 2015

#I=-2arctanx+5/2ln|2x-1| +C#

Explanation:

#I=int (5x^2 – 4x + 7)/((x^2+1)(2x – 1)) dx#

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

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

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

If we compare integrand with expression #T#, we can write:

#2A+C=5#
#2B-A=-4 => A=2B+4#
#-B+C=7 => C=B+7#

#2(2B+4)+B+7=5#
#5B=-10 => B=-2#
#A=0#
#C=5#

#I=int (-2/(x^2+1)+5/(2x-1))dx#

#I=-2arctanx+5/2ln|2x-1| +C#