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

The integral from 0 to 3 of 9/(x^2+9) is (3pi)/4

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

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

\frac{3 π}{4}

Decimal Notation

2.35619 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 9 \cdot \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

= 9 \cdot \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)

= 9 \cdot \frac{1}{3} [arctan (u)]_{0}^{1}

Simplify

9 \cdot \frac{1}{3} [arctan (u)]_{0}^{1} : 3 [arctan (u)]_{0}^{1}

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

Compute the boundaries:

\frac{π}{4}

= 3 \cdot \frac{π}{4}

Simplify

= \frac{3 π}{4}

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What is the integral from 0 to 3 of 9/(x^2+9) ?
The integral from 0 to 3 of 9/(x^2+9) is (3pi)/4