Question

Evaluate: $\int (\frac{1}{\left(lnx\right)}-\frac{1}{(lnx{)}^2})dx$ equals

Answer 1

$\int [\frac{1}{lnx}-\frac{1}{(lnx{)}^2}]dx$

$=\int \frac{1}{ln/x}dx-\int \frac{1}{(lnx{)}^2}dx$

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

$=\frac{x}{lnx}+\int \frac{1}{(lnx{)}^2}dx-\int \frac{1}{(lnx{)}^2}dx$

$=\frac{x}{lnx}+C$

$\therefore \int [\frac{1}{lnx}-\frac{1}{(lnx{)}^2}]dx=\frac{x}{lnx}+C$