What is the integral from 0 to 8/3 of 1/(9x^2+64) ?

The integral from 0 to 8/3 of 1/(9x^2+64) is pi/(96)

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integral from 0 to 8/3 of 1/(9x^2+64)

\int_{0}^{\frac{8}{3}} \frac{1}{9 x ^{2} + 64} d x

\frac{π}{96}

Decimal

0.03272 \dots

Solution steps

\int_{0}^{\frac{8}{3}} \frac{1}{9 x ^{2} + 64} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{24 ( 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}{24} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{24} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{24} \cdot \frac{π}{4}

Simplify

= \frac{π}{96}

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What is the integral from 0 to 8/3 of 1/(9x^2+64) ?
The integral from 0 to 8/3 of 1/(9x^2+64) is pi/(96)