What is the integral from 0 to 8 of 1/(x^2+16) ?

The integral from 0 to 8 of 1/(x^2+16) is 1/4 arctan(2)

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integral from 0 to 8 of 1/(x^2+16)

\int_{0}^{8} \frac{1}{x ^{2} + 16} d x

\frac{1}{4} arctan (2)

Decimal

0.27678 \dots

Solution steps

\int_{0}^{8} \frac{1}{x ^{2} + 16} d x

Apply Integral Substitution

= \int_{0}^{2} \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}^{2} \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}^{2}

Compute the boundaries:

arctan (2)

= \frac{1}{4} arctan (2)

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\frac{\partial}{\partial g} (2 π \frac{l}{g})

What is the integral from 0 to 8 of 1/(x^2+16) ?
The integral from 0 to 8 of 1/(x^2+16) is 1/4 arctan(2)