so im pretty sure its convergent but the answer i get is arctan(6) anyone know what im doing wrong.
the integral is convergent ...
$\displaystyle \lim_{a \to -\infty} \int_a^6 \frac{1}{x^2+1} \, dx$
$\displaystyle \lim_{a \to -\infty} \left[\arctan{x}\right]_a^6$
$\displaystyle \lim_{a \to -\infty} \left[\arctan(6) - \arctan{a}\right] = \arctan(6) - \left(-\frac{\pi}{2}\right) = \arctan(6) + \frac{\pi}{2}$