What is the integral from 0 to 3 of 1/(x^2+16) ?

The integral from 0 to 3 of 1/(x^2+16) is 1/4 arctan(3/4)

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integral from 0 to 3 of 1/(x^2+16)

\int_{0}^{3} \frac{1}{x ^{2} + 16} d x

\frac{1}{4} arctan (\frac{3}{4})

Decimal Notation

0.16087 \dots

Solution steps

\int_{0}^{3} \frac{1}{x ^{2} + 16} d x

Apply Integral Substitution

= \int_{0}^{\frac{3}{4}} \frac{1}{4 ( 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}{4} \cdot \int_{0}^{\frac{3}{4}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{4} [arctan (u)]_{0}^{\frac{3}{4}}

Compute the boundaries:

arctan (\frac{3}{4})

= \frac{1}{4} arctan (\frac{3}{4})

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What is the integral from 0 to 3 of 1/(x^2+16) ?
The integral from 0 to 3 of 1/(x^2+16) is 1/4 arctan(3/4)