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

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

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

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