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

The integral from-1 to 0 of 1/(x^2+1) is pi/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

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

Use the common integral:

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

= [arctan (x)]_{- 1}^{0}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int_{0}^{7} t^{3} - 9 t^{2} + 20 t d t

\int_{0}^{3} x^{2} cos (x) d x

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

\int_{0}^{4} x (x - 2) (x - 4) d x

\int_{0}^{2} 2 t \cdot e^{t^{2}} d t

What is the integral from-1 to 0 of 1/(x^2+1) ?
The integral from-1 to 0 of 1/(x^2+1) is pi/4