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

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{2}

= 2 \cdot \frac{π}{2}

Simplify

= π

\frac{d x}{d t} = 12 - \frac{27 x}{900 - 15 t}

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

\int sec^{2} (4 x) tan^{5} (4 x) d x

\int \frac{ln ( x )}{x ^{9}} d x

f (x) = 6 x^{2} - x^{3}, (1, 5)

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