Mathematics

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

##### SOLUTION
$\int[\dfrac{logx}{(1+logx)^2}]dx$

let $logx=t$
$=>x=e^t$

$dx=e^tdt$
now put the value of t from equation (i) we get,

$\int \dfrac{t}{(1+t)^2}\times e^tdt$

addition and subtraction from $1$

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

we know that,
$f(x)+f'(x)dx=f(x)e^x+c$

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

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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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