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

The integral from 0 to 8 of 1/(64+x^2) is pi/(32)

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

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

\frac{π}{32}

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{8 ( 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}{8} \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}{8} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{8} \cdot \frac{π}{4}

Simplify

= \frac{π}{32}

\frac{d}{d x} (3 x^{2} - 3 x + 4)

n = 0 \sum \infty (- 0.3)^{n}

x^{3} - x^{2} - 4 x - 4, 2 x - 4

\frac{d}{d x} (\frac{8 x ^{2} + 4 x + 4}{x})

5 x^{\frac{3}{2}}

What is the integral from 0 to 8 of 1/(64+x^2) ?
The integral from 0 to 8 of 1/(64+x^2) is pi/(32)