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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- 1}^{1} \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= 2 [arctan (x)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= 2 \cdot \frac{π}{2}

Simplify

= π

\frac{d ^{2}}{d x ^{2}} ((x^{3} + 8 x)^{12})

f (x) = - cos (5 x)

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

\frac{1 - 5 x - x ^{2}}{x ^{2} - 3}

(e^{2 x} + 4) y^{^{'}} = y

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