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

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

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 (1/(x^2+1))

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\frac{\partial}{\partial x} (x^{\frac{3}{4}} y^{\frac{1}{4}})

x \to 0 - lim (\frac{1}{e ^{\frac{1}{x}} - 1})

(x^{3} - x) (x^{2} + 2) (x^{2} + x - 1)

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

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

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