calculate-I-1-2-ln-1-t-t-2-dt-

Question Number 40129 by maxmathsup by imad last updated on 16/Jul/18

c a l c u l a t e I = \int_{1}^{2} \frac{l n (1 + t)}{t^{2}} d t

Answered by maxmathsup by imad last updated on 21/Jul/18

b y p a r t s I = {[- \frac{1}{t} l n (1 + t)]}_{1}^{2} + \int_{1}^{2} \frac{d t}{t (1 + t)}

= l n (2) - \frac{1}{2} l n (3) + \int_{1}^{2} (\frac{1}{t} - \frac{1}{t + 1}) d t

= l n (2) - \frac{1}{2} l n (3) + {[l n ∣ \frac{t}{t + 1} ∣]}_{1}^{2} = l n (2) - \frac{1}{2} l n (3) + l n (\frac{2}{3}) - l n (\frac{1}{2})

I = 3 l n (2) - \frac{3}{2} l n (3)