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

The integral from 0 to 3 of 1/(9+x^2) is pi/(12)

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

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

\frac{π}{12}

Decimal Notation

0.26179 \dots

Solution steps

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

Apply Integral Substitution

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

Compute the boundaries:

\frac{π}{4}

= \frac{1}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{12}

x^{2} - tan (x)

\frac{d y}{d x}, x^{2} - y^{2} = 64

\frac{d}{d x} (sec (\frac{π}{4}))

cos (\frac{x + 1}{x - 1})

F (x) = \frac{3}{1 + x ^{2}}

What is the integral from 0 to 3 of 1/(9+x^2) ?
The integral from 0 to 3 of 1/(9+x^2) is pi/(12)