How do you find the indefinite integral of #(5x^3-42x^2+73x-27)/((x-5)^2(x^2+4))#?

1 Answer
Sep 26, 2015

#2ln|x-5|+3/(x-5)+3/2ln(x^2+4)+1/2arctan(x/2)+C#

Explanation:

#A/(x-5)+B/(x-5)^2+(Cx+D)/(x^2+4)=#

#=(A(x-5)(x^2+4)+B(x^2+4)+(Cx+D)(x-5)^2)/((x-5)^2(x^2+4))=#

#=((Ax-5A)(x^2+4)+Bx^2+4B)/((x-5)^2(x^2+4))+#

#+((Cx+D)(x^2-10x+25))/((x-5)^2(x^2+4))=#

#=(Ax^3-5Ax^2+4Ax-20A+Bx^2+4B)/((x-5)^2(x^2+4))+#

#+(Cx^3-10Cx^2+25Cx+Dx^2-10Dx+25D)/((x-5)^2(x^2+4))=#

#=(x^3(A+C)+x^2(-5A+B-10C+D))/((x-5)^2(x^2+4))+#

#+(x(4A+25C-10D)+(-20A+4B+25D))/((x-5)^2(x^2+4))#

Comparing with #5x^3-42x^2+73x-27# we get system:

#A+C=5#
#-5A+B-10C+D=-42#
#4A+25C-10D=73#
#-20A+4B+25D=-27#

#20A+20C=100#
#-20A+4B-40C+4D=-168#
#-20A-125C+50D=-365#
#-20A+4B+25D=-27#

#4B-20C+4D=-68#
#-105C+50D=-265#
#4B+20C+25D=73#

#40C+21D=141#
#-21C+10D=-53#

#400C+210D=1410#
#441C-210D=1113#

#841C=2523 => C=3#
#D=(141-40C)/21=21/21 => D=1#
#A=5-C => A=2#
#B=-42+5A+10C-D => B=-3#

#int (5x^3-42x^2+73x-27)/((x-5)^2(x^2+4))dx=#

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

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

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