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

Question Number 44472 by abdo.msup.com last updated on 29/Sep/18

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

Commented by maxmathsup by imad last updated on 30/Sep/18

c h a n g e m e n t x t = u g i v e

f (x) = \int_{0}^{\infty} \frac{l n (\frac{u}{x})}{{(1 + u)}^{2}} \frac{d u}{x} = \frac{1}{x} \int_{0}^{\infty} \frac{l n (u) - l n (x)}{{(1 + u)}^{2}} d u

= \frac{1}{x} \int_{0}^{\infty} \frac{l n (u)}{{(1 + u)}^{2}} d u - \frac{l n (x)}{x} \int_{0}^{\infty} \frac{d u}{{(1 + u)}^{2}}

= 0 - \frac{l n (x)}{x} {[- \frac{1}{1 + u}]}_{0}^{+ \infty} = - \frac{l n (x)}{x} \Rightarrow f (x) = - \frac{l n (x)}{x} .