What is the integral from 2 to infinity of 1/(4+x^2) ?

The integral from 2 to infinity of 1/(4+x^2) is pi/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to infinity of 1/(4+x^2)

\int_{2}^{\infty} \frac{1}{4 + x ^{2}} d x

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

\int_{2}^{\infty} \frac{1}{4 + x ^{2}} d x

Compute the indefinite integral:

\int \frac{1}{4 + x ^{2}} d x = \frac{1}{2} arctan (\frac{x}{2}) + C

Compute the boundaries:

\int_{2}^{\infty} \frac{1}{4 + x ^{2}} d x = \frac{π}{4} - \frac{π}{8}

= \frac{π}{4} - \frac{π}{8}

Simplify

= \frac{π}{8}

\int \frac{x ^{2} + x + 1}{( x ^{4} + 2 x ^{2} + 1 ) ^{2}} d x

\int x^{2} arctan (6 x) d x

\int \frac{x ^{3} + 3 x ^{2} + 4}{4 x ^{2} + 2} d x

f (x) = 2 x^{4} + 3 x + 16

y = 3 (x^{2} - 8)^{2}

What is the integral from 2 to infinity of 1/(4+x^2) ?
The integral from 2 to infinity of 1/(4+x^2) is pi/8