How do I find the integral #intt^2/(t+4)dt# ?

1 Answer
Gaurav
Sep 30, 2014

#=t^2/2-4t-24+16ln(t+4)+c#, where #c# is a constant

Explanation :

#=intt^2/(t+4)dt#

let's #t+4=u#, then #dt=du#

#=int(u-4)^2/udu#

#=int(u^2-8u+16)/udu#

#=int(u^2/u-8u/u+16/u)du#

#=int(u-8+16/u)du#

#=intudu-int8du+int16/udu#

#=u^2/2-8u+16lnu+c#, where #c# is a constant

Substituting #u# back yields,

#=(t+4)^2/2-8(t+4)+16ln(t+4)+c#, where #c# is a constant

Simplifying further,

#=t^2/2+8+4t-8t-32+16ln(t+4)+c#, where #c# is a constant

#=t^2/2-4t-24+16ln(t+4)+c#, where #c# is a constant

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