What is the integral from-3 to 3 of 1/(1+x^2) ?

The integral from-3 to 3 of 1/(1+x^2) is 2arctan(3)

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integral from-3 to 3 of 1/(1+x^2)

\int_{- 3}^{3} \frac{1}{1 + x ^{2}} d x

2 arctan (3)

Decimal Notation

2.49809 \dots

Solution steps

\int_{- 3}^{3} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= [arctan (x)]_{- 3}^{3}

Compute the boundaries:

2 arctan (3)

= 2 arctan (3)

\frac{d y}{d x} = x^{8} y^{- 7}

t \cdot \frac{d y}{d t} = t^{3} + 9 t^{3} y

\frac{d}{d x} ((x^{2} + x)^{2} sin^{3} (2 x))

x \to 0 lim (ln (x + e^{2 x}))

\int \frac{1}{x ^{2} + 6 x + 9} d x

What is the integral from-3 to 3 of 1/(1+x^2) ?
The integral from-3 to 3 of 1/(1+x^2) is 2arctan(3)