How do you integrate #int tln(t+1)# by parts?

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Answer 1 · 2016-11-18T15:56:50

#I=1/2t^2ln(t+1)-1/4t^2-1/2t+1/2ln|t+1|+C#

#intu(dv)/(dt)dt=uv-intv(du)/(dt)dt#

note function not defined for #t=-1#

for

#int(tln(t+1))dt#

note function not defined for #t=-1#

#u=ln(t+1)=>(du)/(dt)=1/(t+1)#

#(dv)/(dt)=t=>v=1/2t^2#

#:. intu(dv)/(dt)dt=1/2t^2ln(t+1)-1/2intt^2/(t+1)dt#

divide the improper fraction out

#:. intu(dv)/(dt)dt=1/2t^2ln(t+1)-1/2int((t-1)+1/(t+1))dt#

#:. intu(dv)/(dt)dt=1/2t^2ln(t+1)-1/2(1/2t^2-t+ln|t+1|)+C#

#I=1/2t^2ln(t+1)-1/4t^2-1/2t+1/2ln|t+1|+C#