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

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

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

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

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

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

\int_{0}^{3} (x^{2} - x + 1) e^{2 x} d x

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

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\int_{\frac{π}{6}}^{\frac{π}{2}} x csc^{2} (x) d x

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