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

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

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

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

\frac{π}{2} - arctan (2)

Decimal

0.46364 \dots

Solution steps

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

= \frac{π}{2} - arctan (2)

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\int_{- 1}^{2} (x^{3} - x^{2} - 2 x) d x

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\int_{0}^{2} e^{- 2 x + 1} d x

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