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

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

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

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

2 arctan (2)

Decimal Notation

2.21429 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

2 arctan (2)

= 2 arctan (2)

f (x) = \frac{2 x}{2 e ^{9 x}}

\frac{d}{d x} (ln (3 x - 5))

y = \frac{( x - 2 ) ( x ^{2} + 2 x + 4 )}{x ^{3}}

x^{2} + y^{2}

\int_{0}^{1} 2 π (y) (y - y^{3}) d y

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