How do you integrate #((3x^2)-25x+43)dx/((2x+1)((x-2)^2))#?
1 Answer
Explanation:
Lets rewrite integrand
#(3x^2-25x+43)/((2x+1)(x-2)^2)=A/(2x+1)+B/(x-2)+C/(x-2)^2=#
#=(A(x-2)^2+B(2x+1)(x-2)+C(2x+1))/((2x+1)(x-2)^2)=#
#=(A(x^2-4x+4)+B(2x^2-3x-2)+C(2x+1))/((2x+1)(x-2)^2)=#
#=(Ax^2-4Ax+4A+2Bx^2-3Bx-2B+2Cx+C)/((2x+1)(x-2)^2)=#
#=(x^2(A+2B)+x(-4A-3B+2C)+(4A-2B+C))/((2x+1)(x-2)^2)#
Comparing with integrand coefficients we get system of linear equations with 3 unknowns:
#A+2B=3 => A=3-2B#
#-4A-3B+2C=-25#
#4A-2B+C=43#
From 1st eq insert into the 2nd and 3rd:
#-4(3-2B)-3B+2C=-25#
#4(3-2B)-2B+C=43#
#-12+8B-3B+2C=-25#
#12-8B-2B+C=43#
#5B+2C=-13#
#-10B+C=31#
#5B+2C=-13#
#-20B+2C=62#
#25B=-75 => B=-3#
#C=31+10B=31-30=1 => C=1#
#A=3-2B=3+6=9 => A=9#
So, the integral is broken apart:
#int9/(2x+1)dx+int(-3)/(x-2)dx+int1/(x-2)^2dx=I#
#2x+1=t, 2dx=dt, dx=dt/2,#
#x-2=u, dx=du,#
#x-2=m, dx=dm#
#I=int9/tdt/2+int(-3)/udu+int1/m^2dm=#
#=9/2intdt/t-3int(du)/u+intm^-2dm=#
#=9/2lnabs(t)-3lnabs(u)+m^-1/-1+C=#
#=9/2lnabs(2x+1)-3lnabs(x-2)-1/(x-2)+C#