Integeration of u^3/1+u ?

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Answer 1 · 2018-03-03T17:35:17

#(1+u)^3-3/2(1+u)2+3(1+u)-ln|(1+u)|+c#

here is one method

#I=int(u^3)/(1+u)du#

using the substitution

#t=1+u=>dt=du#

#I=int(t-1)^3/tdt#

#=int(t^3-3t^2+3t-1)/tdt#

#=int(t^2-3t+3-1/t)dt#

#=t^3-3/2t^2+3t-ln|t|+c#

substitute back for #t#

#=(1+u)^3-3/2(1+u)2+3(1+u)-ln|(1+u)|+c#

which can be further simplified if desired