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

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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

\int e^{- 0.4 x} d x

x \to - 2 lim (\frac{x + 2}{x ^{3} + 8})

z \to 2 lim (6 z - 1)

7 y^{^{''}} + 2 y^{^{'}} = 0

n = 1 \sum \infty \frac{- 1}{n}

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