find-the-value-of-0-1-ln-t-1-t-2-dt-

Question Number 40154 by maxmathsup by imad last updated on 16/Jul/18

f i n d t h e v a l u e o f \int_{0}^{1} \frac{l n (t)}{{(1 + t)}^{2}} d t

Commented by maxmathsup by imad last updated on 16/Jul/18

l e t I = \int_{0}^{1} \frac{l n (t)}{{(1 + t)}^{2}} d t l e t i n t e g r a t e b y p a r t s u^{^{'}} = \frac{1}{{(1 + t)}^{2}} a n d v = l n (t)

I = {[(1 - \frac{1}{1 + t}) l n (t)]}_{0}^{1} - \int_{0}^{1} (1 - \frac{1}{1 + t}) \frac{d t}{t}

= - \int_{0}^{1} \frac{d t}{1 + t} = - {[l n ∣ 1 + t ∣]}_{0}^{1} = - l n (2)

I = - l n (2) .

Answered by ajfour last updated on 16/Jul/18

I = \int_{0}^{1} \frac{\ln t}{{(1 + t)}^{2}}

= - \frac{\ln t}{(1 + t)} ∣_{0}^{1} + \int_{0}^{1} \frac{d t}{t (1 + t)}

= \lim_{t \to 0} [(\frac{\ln t}{1 + t}) - \ln (\frac{t}{1 + t})] - \ln 2

= \lim_{t \to 0} [\ln \frac{(1 + t) t^{1 + t}}{t}] - \ln 2

= \lim_{t \to 0} [\ln (1 + t) t^{t}] - \ln 2

= \underset{t \to 0}{\lim l n} (1 + t) + \lim_{t \to 0} \frac{\ln t}{(1 / t)} - \ln 2

= 0 + \lim_{t \to 0} (\frac{1 / t}{- 1 / t^{2}}) - \ln 2

I = - \ln 2 .

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