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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Decimal

0.67551 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

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