How do you find the indefinite integral of #int (x^2-6x-20)/(x+5)#?

1 Answer
Andrea S.
Nov 30, 2016

#int frac (x^2-6x-20) (x+5) dx = (x+5)^2/2 -16(x+5) +35 ln (x+5)#

Explanation:

Substitute #x=t-5#

#int frac (x^2-6x-20) (x+5) dx = int (dt)/t((t-5)^2 -6(t-5) -20) =#

#= int (dt)/t(t^2-10t +25 -6t+30 -20) =#

#= int (dt)/t(t^2-16t +35) = int t dt -16 int dt +35 int (dt)/t = #

# = (t^2)/2 -16t +35 ln(t) = (x+5)^2/2 -16(x+5) +35 ln (x+5)#

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