find-f-x-1-2-ln-1-xt-t-2-dt-with-x-gt-0-

Question Number 57228 by maxmathsup by imad last updated on 31/Mar/19

f i n d f (x) = \int_{1}^{2} \frac{l n (1 + x t)}{t^{2}} d t w i t h x > 0

Commented by maxmathsup by imad last updated on 12/Apr/19

b y p a r t s u^{^{'}} = \frac{1}{t^{2}} a n d v = l n (1 + x t) \Rightarrow f (x) = {[- \frac{1}{t} l n (1 + x t)]}_{1}^{2}

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

\int_{1}^{2} \frac{x}{t (x t + 1)} d t =_{x t = u} \int_{x}^{2 x} \frac{x}{\frac{u}{x} (u + 1)} \frac{d u}{x} = \int_{x}^{2 x} \frac{x}{u (u + 1)} d t

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

f (x) = l n (1 + x) - \frac{1}{2} l n (1 + 2 x) + x l n (\frac{2 x}{2 x + 1}) - x l n (\frac{x}{x + 1}) .