What is the integral from 0 to1 of (4/(t^2+1))dt?

${\int}_{0}^{1} \left(\frac{4}{{t}^{2} + 1}\right) \mathrm{dt} = \pi$
${\int}_{0}^{1} \left(\frac{4}{{t}^{2} + 1}\right) \mathrm{dt} = 4 {\int}_{0}^{1} \left(\frac{1}{1 + {t}^{2}}\right) \mathrm{dt}$
$= 4 {\left[\arctan t\right]}_{0}^{1} = 4 \left(\arctan 1 - \arctan 0\right) = 4 \left(\frac{\pi}{4} - 0\right) = \pi$