WEBVTT

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Here is the problem: 1/(x squared plus 1).

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Evaluate the improper integral of 1 / (x squared plus 1) from 0 to inf.

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This is an improper integral. It is improper because a limit of integration is infinite. It is defined as a limit, and it converges only if every limit is finite.

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Inverse tangent form: the integral of du/(a squared plus u squared) equals (1/a) times arctan(u/a).

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Antiderivative.

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Evaluate the piece from 0 to inf.

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Add the pieces. Every limit is finite, so the integral converges.

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So the answer is Converges: pi/2 approximately 1.570796327.

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An improper integral replaces the problem point with a variable, integrates normally, and then takes a limit. A finite limit means the infinite region still has finite area.

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Antiderivative verified by differentiation, which checks out.

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Each limit of the antiderivative was computed and checked numerically by the limit engine, which checks out.

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Truncated numerical estimate (stopping 10 to the power minus 6 from asymptotes and at plus or minus 1000 for infinite limits): 1.5698.