Wolfram Alpha gives the indefinite integral $\displaystyle\int \ln\left(1+\frac{1}{t^2}\right)\,\mathrm dt = t\ln\left(1+\frac{1}{t^2}\right) + 2\arctan t$.
It is easy to see that integration by parts will work.
Set $u = \ln\left(1+\dfrac{1}{t^2}\right)$ and $\mathrm dv =\mathrm dt$, we have $\mathrm du = \dfrac{1}{1+\frac{1}{t^2}} \cdot \dfrac{-2}{t^3}\,\mathrm dt = \dfrac{-2}{t^3+t}$ and $v = t$.
Thus, $\displaystyle\int \ln\left(1+\frac{1}{t^2}\right)\,dt = t\ln\left(1+\frac{1}{t^2}\right)- \int t \dfrac{-2}{t^3+t}\,\mathrm dt$ $= t\ln\left(1+\frac{1}{t^2}\right) + \int \dfrac{2}{t^2+1}\,\mathrm dt$ $= t\ln\left(1+\dfrac{1}{t^2}\right) + 2\arctan t$.
Since $\displaystyle\lim_{t\to 0^+}t\ln\left(1+\dfrac{1}{t^2}\right) + 2\arctan t = 0$ and $\displaystyle\lim_{t\to \infty}t\ln\left(1+\dfrac{1}{t^2}\right) + 2\arctan t = \pi$, we have
$\displaystyle\int_{0}^{\infty} \ln\left(1+\frac{1}{t^2}\right)\,\mathrm dt = \pi$.