$\displaystyle\int {\dfrac{{\ln \left( {1 + {x}} \right)}}{{1 + {x}}}} dx\,equals$

Given

$\displaystyle \int \dfrac {\ln (1+x)}{1+x}dx$

let $1+x=t$

$\implies dx=dt$

$\implies \displaystyle \int \dfrac {\ln t}{t}dt$

Let $\ln t=p$

$\implies \dfrac 1tdt =dp$

$\implies \displaystyle \int pdp$

$\implies \dfrac{p^2}{2}$

$\implies \dfrac {(\ln t)^2}2$

$\implies \dfrac {(\ln (1+x))^2}2$