WEBVTT

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Here is the problem: integral of 1/(x squared plus 9).

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Find the indefinite integral (antiderivative) of 1 / (x squared plus 9) with respect to x.

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Start with the integral.

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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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Simplify and add the constant of integration.

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So the answer is atan(x / 3) / 3 plus C.

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An antiderivative F satisfies F prime of x equals f(x). Because the derivative of any constant is 0, every antiderivative differs by a constant C.

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Differentiated the answer and compared it with the integrand at 8 sample points, which checks out.