Question

$I=\int \frac{\log x}{{\left(1+\log x\right)}^2}dx$

Answer 1

$\int \left[\frac{logx}{(1+logx{)}^2}\right]dx$

let $logx=t$

$=>x=e^t$

$dx=e^{t}dt$

now put the value of t from equation (i) we get,

$\int \frac{t}{(1+t{)}^2}\times e^{t}dt$

addition and subtraction from $1$

$\int [\frac{t+1-1}{(1+t{)}^2}\times e^t]dt$

$\int [\frac{1}{1+t}]-\frac{1}{(1+t{)}^2}]dt$

we know that,

$f\left(x\right)+f^{\prime }\left(x\right)dx=f\left(x\right)e^x+c$

$=>\frac{e^t}{1+t}+c$

put the value of $t$ anf $e^t$,we get

$=>\frac{x}{1+logx}+c$