Answer
The integral converges and its value is $\pi/2$.
Work Step by Step
We have
$$
\int_{-\infty}^{0} \frac{d x}{x^2+1}=\tan^{-1}x|_{-\infty}^0\\
=\tan^{-1}0-\tan^{-1}(-\infty)=0-(-\pi/2)=\pi/2.
$$
Hence, the integral converges and its value is $\pi/2$.
