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

The integral from 0 to k of 1/(x^2+1) is arctan(k)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to k of 1/(x^2+1)

\int_{0}^{k} \frac{1}{x ^{2} + 1} d x

arctan (k)

Solution steps

\int_{0}^{k} \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= [arctan (x)]_{0}^{k}

Compute the boundaries:

arctan (k)

= arctan (k)

\int \frac{e ^{2 x}}{2} d x

n = 0 \sum \infty \frac{n ^{3}}{n !}

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 7 x - y}

\frac{d}{d x} (\frac{- 2 x}{( x ^{2} - 1 ) ^{2}})

\int \frac{( 2 x ^{3} - x )}{x ^{4} - x ^{2} - 25} d x

What is the integral from 0 to k of 1/(x^2+1) ?
The integral from 0 to k of 1/(x^2+1) is arctan(k)