What is the integral from a to infinity of 1/(x^2+1) ?

The integral from a to infinity of 1/(x^2+1) is pi/2-arctan(a)

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integral from a to infinity of 1/(x^2+1)

\int_{a}^{\infty} \frac{1}{x ^{2} + 1} d x

\frac{π}{2} - arctan (a)

Solution steps

\int_{a}^{\infty} \frac{1}{x ^{2} + 1} d x

Compute the indefinite integral:

\int \frac{1}{x ^{2} + 1} d x = arctan (x) + C

Compute the boundaries:

\int_{a}^{\infty} \frac{1}{x ^{2} + 1} d x = \frac{π}{2} - arctan (a)

= \frac{π}{2} - arctan (a)

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What is the integral from a to infinity of 1/(x^2+1) ?
The integral from a to infinity of 1/(x^2+1) is pi/2-arctan(a)