Question

$\int [\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2}]dx=x{\left(\log x\right)}^{-1}.$ If this is true enter 1, else enter 0.

Answer 1

Let $I=\int [\frac{1}{\log x}-\frac{1}{{\left(\log x\right)}^2}]dx$

$I=I_1-I_2$

$I_1=\int \frac{1}{\log x}dx=x.\frac{1}{\log x}-\int x.{(-\frac{1}{\log x})}^2.\frac{1}{x}dx$

$I_1=x.\frac{1}{\log x}+\int {\left(\frac{1}{\log x}\right)}^{2}dx=x.\frac{1}{\log x}+I_2$

or $I=I_1-I_2=x{\left(\log x\right)}^{-1}.$