integrate-by-use-a-partial-friction-lnx-1-x-2-

Question Number 43191 by MASANJA J last updated on 08/Sep/18

i n t e g r a t e b y u s e a p a r t i a l f r i c t i o n

\int \frac{l n x}{{(1 + x)}^{2}}

Commented by mondodotto@gmail.com last updated on 09/Sep/18

did you mean by parts ?

Commented by maxmathsup by imad last updated on 08/Sep/18

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

I = - \frac{l n (x)}{1 + x} + \int \frac{1}{(1 + x) x} d x b u t \int \frac{d x}{x (1 + x)} = \int (\frac{1}{x} - \frac{1}{x + 1}) d x = l n ∣ \frac{x}{x + 1} ∣ + c \Rightarrow

I = l n ∣ \frac{x}{x + 1} ∣ - \frac{l n (x)}{1 + x} + c .

Commented by maxmathsup by imad last updated on 09/Sep/18

i n t e g r a t i o n b y p a r t s .

Answered by alex041103 last updated on 08/Sep/18

f i r s t w e u s e I B P :

\int \frac{l n x}{{(1 + x)}^{2}} d x = \int - l n x d (\frac{1}{1 + x}) =

= - (\frac{l n x}{1 + x} - \int \frac{d (l n x)}{1 + x}) =

= \int \frac{d x}{x (1 + x)} - \frac{l n (x)}{1 + x}

\frac{1}{x (1 + x)} = \frac{1}{x} - \frac{1}{1 + x}

\Rightarrow \int \frac{d x}{x (1 + x)} = \int \frac{d x}{x} - \int \frac{d (x + 1)}{1 + x} =

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

\Rightarrow \int \frac{l n x}{{(1 + x)}^{2}} d x = l n (\frac{x}{1 + x}) - \frac{l n (x)}{1 + x} + C