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

The integral from-3 to 3 of 1/(x^2+9) is pi/6

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

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

\frac{π}{6}

Solution steps

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

Apply Integral Substitution

= \int_{- 1}^{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_{- 1}^{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)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{1}{3} \cdot \frac{π}{2}

Simplify

= \frac{π}{6}

10 cos (t)

\frac{d}{d x} ((3 x - 1)^{4} (7 x^{2} - 4)^{- 3})

\int 4 x (3 x - 1)^{5} d x

x^{2} + 2 x y - y^{2} + x = 20, (3, 4)

(3 x + 6)^{\frac{1}{5}}, at x = 2

What is the integral from-3 to 3 of 1/(x^2+9) ?
The integral from-3 to 3 of 1/(x^2+9) is pi/6