Question

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

Answer 1

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

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

$=\left(\frac{x}{\log x}\right)+$$\int \left(\frac{1}{(\log x{)}^2}\right).dx$$-\int \left(\frac{1}{(\log x{)}^2}\right).dx$ [ Using method of by parts]

$=\left(\frac{x}{\log x}\right)+c$ [ Where $c$ is integrating constant]