Question

How do you evaluate the integral `int 1/(x(1+(lnx)^2)`?

Answer 1

`int1/(x(1+(lnx)^2))dx=arctan(lnx)+C`

Explanation:

Let `u=lnx`. This implies that `du=1/xdx`.

`int1/(x(1+(lnx)^2))dx=int1/(1+(lnx)^2)(1/xdx)=int1/(1+u^2)du`

This is the arctangent integral:

`=arctan(u)+C=arctan(lnx)+C`