# Calc

Find the integral from -∞ to -1 of 1/(1+x²) dx.

I started working the problem out, and so far I got the lim as z→-∞ of [arctan (-1) - arctan (z)].

I'm a little bit confused exactly what is means when you plug -∞ into z, and how to solve the problem from here.

Thank you!

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1. The indefinite integral is arctan x, as you have already indicated.

arctan(-infinity), the angle that has a tangent of -infinity, is (3/2)*pi

arctan(-1) = (7/8)*(2 pi) = (7/4)pi

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posted by drwls

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