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

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

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

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

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