What is the integral from-1 to 1 of (6/(x^2+1)) ?

The integral from-1 to 1 of (6/(x^2+1)) is 3pi

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integral from-1 to 1 of (6/(x^2+1))

\int_{- 1}^{1} (\frac{6}{x ^{2} + 1}) d x

3 π

Decimal

9.42477 \dots

Solution steps

\int_{- 1}^{1} \frac{6}{x ^{2} + 1} d x

Take the constant out:

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

= 6 \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)

= 6 [arctan (x)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= 6 \cdot \frac{π}{2}

Simplify

= 3 π

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What is the integral from-1 to 1 of (6/(x^2+1)) ?
The integral from-1 to 1 of (6/(x^2+1)) is 3pi