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

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

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

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

\frac{π}{2} - 2 arctan (\frac{1}{2})

Decimal

1.35102 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{π}{2} - 2 arctan (\frac{1}{2})

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