What is the integral from 0 to infinity of 2/(1+t^2) ?

The integral from 0 to infinity of 2/(1+t^2) is pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of 2/(1+t^2)

\int_{0}^{\infty} \frac{2}{1 + t ^{2}} d t

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{0}^{\infty} \frac{2}{1 + t ^{2}} d t

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{\infty} \frac{2}{1 + t ^{2}} d t = π - 0

= π - 0

Simplify

= π

\int_{0}^{\frac{1}{2}} e^{- x^{3}} d x

\int_{1}^{3} ln (9 x) d x

\int_{- 2}^{0} \frac{1}{x ^{4}} d x

\int_{1}^{e} 8 ln (x) d x

\int_{0}^{2} y + x e^{y} d x

What is the integral from 0 to infinity of 2/(1+t^2) ?
The integral from 0 to infinity of 2/(1+t^2) is pi