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

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

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

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

- arctan (2)

Decimal

- 1.10714 \dots

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= - [arctan (x)]_{0}^{2}

Compute the boundaries:

arctan (2)

= - arctan (2)

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

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

\int_{- 5}^{8} (10 - 5 x) d x

\int_{- 1}^{0} - 2 x^{3} (- x^{4} + 2)^{4} d x

\int_{0}^{3} 5 x (x - 4) d x

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