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

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

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integral from-3 to 3 of 6/(x^2+9)

\int_{- 3}^{3} \frac{6}{x ^{2} + 9} d x

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{- 3}^{3} \frac{6}{x ^{2} + 9} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} + 9} d x

Apply Integral Substitution

= 6 \cdot \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

= 6 \cdot \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)

= 6 \cdot \frac{1}{3} [arctan (u)]_{- 1}^{1}

Simplify

6 \cdot \frac{1}{3} [arctan (u)]_{- 1}^{1} : 2 [arctan (u)]_{- 1}^{1}

= 2 [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= 2 \cdot \frac{π}{2}

Simplify

= π

\int x^{9} \cdot e^{x^{10} - 9} d x

y = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

n = 0 \sum \infty e^{- \frac{9 n}{2}}

\int u^{- 3} d u

\frac{d}{d t} (t^{\frac{3}{2}})

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