∫(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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Originally Posted by mike789 ∫(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? Notations Gives us
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