Hey,

I am on a problem that involves integrating a arctan:

$\displaystyle \int{\arctan{\left(\frac{1}{x}\right)}}dx$

I think what is throwing me off is the $\displaystyle \frac{1}{x}$

I set:

$\displaystyle u=\arctan{\left(\frac{1}{x}\right)}$

$\displaystyle dv=dx$

so:

$\displaystyle du=\frac{1}{1+\left(\frac{1}{x}\right)^2}$

$\displaystyle v=x$

and I get:

$\displaystyle x\arctan{\left(\frac{1}{x}\right)}-\int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx$

I think $\displaystyle \int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx$ is the problem for me. No matter how I work it, I can't seem to get the answer (maple):

$\displaystyle x\arctan{\left(\frac{1}{x}\right)}+\frac{1}{2}\ln{ \left(1+\frac{1}{x^2}\right)}-\ln{\left(\frac{1}{x}\right)}$

Any ideas?