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

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

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

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

\frac{π}{2} - arctan (s)

Solution steps

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

= \frac{π}{2} - arctan (s)

\int_{0}^{π} sin (x) d x

\int_{0}^{1} y sin (y) d y

\int_{0}^{1} arctan (1 - x) d x

\int_{- 1}^{- 2} \frac{e ^{x}}{1 - e ^{x}} d x

\int_{- 1}^{1} \frac{5}{x ^{2} + 1} d x

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