I-0-1-lnt-1-t-2-dt-Helpe-me-please-

Question Number 120383 by SOMEDAVONG last updated on 31/Oct/20

I / . \int_{0}^{1} \frac{lnt}{{(1 + t)}^{2}} dt = ?

(Helpe me please)

Answered by Dwaipayan Shikari last updated on 31/Oct/20

\int_{0}^{1} \frac{l o g t}{{(1 + t)}^{2}} d t

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

= {[l o g (\frac{t}{t + 1})]}_{0}^{1} = - l o g (2)

Commented by SOMEDAVONG last updated on 31/Oct/20

(Thank you so much)

Answered by Olaf last updated on 31/Oct/20

I (t) = \int \frac{\ln t}{{(1 + t)}^{2}} d t

By parts :

I (t) = \ln t (- \frac{1}{1 + t}) - \int \frac{1}{t} (- \frac{1}{1 + t}) d t

I (t) = - \frac{\ln t}{1 + t} + \int (\frac{1}{t} - \frac{1}{1 + t}) d t

I (t) = - \frac{\ln t}{1 + t} + \ln t - \ln (t + 1)

I (t) = \ln t (1 - \frac{1}{1 + t}) - \ln (t + 1)

I (t) = \frac{t \ln t}{1 + t} - \ln (t + 1)

I = I (1) - I (0) = (0 - \ln 2) - (0 - 0)

I = - \ln 2

Commented by SOMEDAVONG last updated on 31/Oct/20

(Thank you so much sir)

Answered by Lordose last updated on 31/Oct/20

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

IBP

I = ∣ \frac{lnt}{- (1 + t)} ∣_{0}^{1} + \int_{0}^{1} \frac{1}{t (1 + t)} dt

I =∣ \frac{lnt}{- (1 + t)} ∣_{0}^{1} + \int_{0}^{1} (\frac{1}{t} - \frac{1}{1 + t}) dt

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

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

I = - \ln (2)

Commented by SOMEDAVONG last updated on 31/Oct/20

(Thank you so much)