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

Question

1382 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-08T07:06:58

# I=-(9x+11)/(3(x+2)^3)+c'.#

I feel that there is no need to use the Method of Partial Fractions,

because, without it, the Problem can be easily worked out, as shown

below:

#I=int(6x+5)/(x+2)^4dx=int{(6x+12)-7}/(x+2)^4dx,#

#=int(6(x+2))/(x+2)^4dx-7int1/(x+2)^4dx,#

#=6int(x+2)^-3dx-7int(x+2)^-4dx.#

We know that,

#intf(x)dx=F(x)+C rArr intf(ax+b)dx=1/aF(ax+b)+C, ane0.#

Since #intx^-3dx=x^(-3+1)/(-3+1)=-1/(2x^2)+c,# we have,

#I=6{-1/(2(x+2)^2)}-7{(x+2)^(-4+1)/(-4+1)},#

#=7/(3(x+2)^3)-3/(x+2)^2={7-9(x+2)}/(3(x+2)^3).#

#:. I=-(9x+11)/(3(x+2)^3)+c'.#

Enjoy Maths.!