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

1 Answer
Nov 18, 2016

I=1/2t^2ln(t+1)-1/4t^2-1/2t+1/2ln|t+1|+CI=12t2ln(t+1)14t212t+12ln|t+1|+C

Explanation:

integration by parts formula

intu(dv)/(dt)dt=uv-intv(du)/(dt)dtudvdtdt=uvvdudtdt

note function not defined for t=-1t=1

for

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

note function not defined for t=-1t=1

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

(dv)/(dt)=t=>v=1/2t^2dvdt=tv=12t2

:. 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