Question

How do you find the integral `ln(x)/(x^2)`?

Answer 1

`-lnx /x -1/x +C`

Explanation:

Integration by parts can be done in this case, `int (ln x)/x^2 dx` =`int lnx*d/dx(-1/x) dx`= `-lnx /x -int d/dx (lnx)*(-1)/x dx+C`= `-lnx /x+int 1/x^2 dx +C`= `-lnx /x -1/x +C`