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
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