Mathematics

$$\displaystyle \int {\left\{ {\dfrac{1}{{\log x}} - \dfrac{1}{{{{\left( {\log x} \right)}^2}}}} \right\}dx} $$


SOLUTION
$$\int \left\{\dfrac{1}{\log\,x}-\dfrac{1}{(\log\,x)^2}\right\}dx$$
Now, take $$\log\,x=t$$
$$\Rightarrow$$  $$x=e^t$$
$$\Rightarrow$$  $$dx=e^tdt$$
The given integral becomes,
$$\int\left(\dfrac{1}{t}\right)-\left(\dfrac{1}{t^2}\right)dt$$
$$\Rightarrow$$  $$\int e^t\left(\dfrac{1}{t}-\dfrac{1}{t^2}\right)dt$$
Now, we have a formula,
$$\int e^x(f(x)+f'(x))dx=e^xf(x)$$         ------ ( 1 )
So here, left $$f(x)=\dfrac{1}{t}\Rightarrow \,\,f'(x)=\dfrac{-1}{t^2}$$
So, it is of form ( 1 )
$$\therefore$$  The given integral $$=e^xf(x)=e^t(1/t)+c$$
Substituting value of $$t$$ we get,
$$\Rightarrow$$  $$e^{\log\,x}(\dfrac{1}{\log\,x})+c$$
$$\Rightarrow$$  $$\dfrac{x}{\log\,x}+c$$
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Subjective Medium Published on 17th 09, 2020
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