Question

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

Answer 1

$\int \{\frac{1}{\log x}-\frac{1}{(\log x{)}^2}\}dx$

Now, take $\log x=t$

$\Rightarrow$ $x=e^t$

$\Rightarrow$ $dx=e^{t}dt$

The given integral becomes,

$\int \left(\frac{1}{t}\right)-\left(\frac{1}{t^2}\right)dt$

$\Rightarrow$ $\int e^t(\frac{1}{t}-\frac{1}{t^2})dt$

Now, we have a formula,

$\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)$ ------ ( 1 )

So here, left $f\left(x\right)=\frac{1}{t}\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{t^2}$

So, it is of form ( 1 )

$\therefore$ The given integral $=e^{x}f\left(x\right)=e^t\left(1/t\right)+c$

Substituting value of $t$ we get,

$\Rightarrow$ $e^{\log x}\left(\frac{1}{\log x}\right)+c$

$\Rightarrow$ $\frac{x}{\log x}+c$