How do you integrate #int (6x+5) / (x+2) ^4# using partial fractions?

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Answer 1 · 2016-03-30T18:53:03

#int(6x+5)/(x+2)^4 dx = int (6/(x+2)^3-7/(x+2)^4)dx = -2/(x+2)^2+7/(3(x+2)^3) + C#

#(6x+5)/(x+2)^4 = A/((x+2)) + B/(x+2)^2 +C/(x+2)^3 +D/(x+2)^4 #

#A(x+2)^3 + B(x+2)^2 +C(x+2) + D = 6x+5#

#Ax^3 = 0x^3 rArr A=0#

#Bx^2 = 0x^2 rArr B=0#

#Cx = 6x rArr C=6#

#2C+D=5 and C=6 rArr D=-7#

Now evaluate #int(6(x+2)^-3 - 7(x+2)^-4) dx# to get

#-2/(x+2)^2+7/(3(x+2)^3) + C#