Answer
$\frac{\pi}{2\sqrt{3}}+\ln{2}-\frac{\pi}{4}-\frac{\ln{2}}{2}$
Work Step by Step
First find the indefinite integral:
Let $u=\arctan{\frac{1}{x}}$ and $dv=1 \ dx$. Then $du=\frac{-1}{1+x^2}$ and $v=x$. Apply integration by parts:
$$\int \arctan{\frac{1}{x}} \ dx = x\arctan\frac{1}{x} - \int \frac{x}{1+x^2} \ dx$$
To evaluate the integral, we need to do a substitution. Let $k=x^2$. Then $\frac{1}{2} dk=x \ dx$
$$ \int \frac{x}{1+x^2} \ dx = -\frac{1}{2} \int \frac{1}{1+k} \ dk$$
$$=-\frac{1}{2}\ln\vert1+k\vert$$
$$=-\frac{1}{2}\ln\vert1+x^2\vert$$
So then
$$x\arctan\frac{1}{x} - \int \frac{x}{1+x^2} \ dx =x\arctan\frac{1}{x} + \frac{1}{2}\ln\vert1+x^2\vert$$
Now evaluate for $x=1,\sqrt{3}$:
$$x\arctan\frac{1}{x} + \frac{1}{2}\ln\vert1+x^2\vert \Big]^\sqrt3_1=\frac{\pi}{2\sqrt{3}}+\ln{2}-\frac{\pi}{4}-\frac{\ln{2}}{2}$$
