What is the integral from-16 to 16 of 1/(256+x^2) ?

The integral from-16 to 16 of 1/(256+x^2) is pi/(32)

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integral from-16 to 16 of 1/(256+x^2)

\int_{- 16}^{16} \frac{1}{256 + x ^{2}} d x

\frac{π}{32}

Decimal Notation

0.09817 \dots

Solution steps

\int_{- 16}^{16} \frac{1}{256 + x ^{2}} d x

Apply Integral Substitution

= \int_{- 1}^{1} \frac{1}{16 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{16} [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{1}{16} \cdot \frac{π}{2}

Simplify

= \frac{π}{32}

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What is the integral from-16 to 16 of 1/(256+x^2) ?
The integral from-16 to 16 of 1/(256+x^2) is pi/(32)