∫(t-1)ln(t)dt

U=lnt dv/dx = t-1

Du/dx = 1/t V = ((t^2)/2)-t

=lnt((t^2/2)-t)- ∫((t^2/2)-t)*1/t

After doing the calculations I got:

((t^2/2)-t)lnt-(t^2/4)+t/2

The correct answer is

((t^2/2)-t)lnt-(t^2/4)+t

Any help..plz?

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- January 14th 2011, 11:44 AMmike789Integrate:∫(t-1)ln(t)dt
∫(t-1)ln(t)dt

U=lnt dv/dx = t-1

Du/dx = 1/t V = ((t^2)/2)-t

=lnt((t^2/2)-t)- ∫((t^2/2)-t)*1/t

After doing the calculations I got:

((t^2/2)-t)lnt-(t^2/4)+t/2

The correct answer is

((t^2/2)-t)lnt-(t^2/4)+t

Any help..plz? - January 14th 2011, 11:56 AMPranas