What is the integral from 0 to 4 of (36)/(x^2+9) ?

The integral from 0 to 4 of (36)/(x^2+9) is 12arctan(4/3)

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integral from 0 to 4 of (36)/(x^2+9)

\int_{0}^{4} \frac{36}{x ^{2} + 9} d x

12 arctan (\frac{4}{3})

Decimal Notation

11.12754 \dots

Solution steps

\int_{0}^{4} \frac{36}{x ^{2} + 9} d x

Take the constant out:

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

= 36 \cdot \int_{0}^{4} \frac{1}{x ^{2} + 9} d x

Apply Integral Substitution

= 36 \cdot \int_{0}^{\frac{4}{3}} \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

= 36 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{4}{3}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 36 \cdot \frac{1}{3} [arctan (u)]_{0}^{\frac{4}{3}}

Simplify

36 \cdot \frac{1}{3} [arctan (u)]_{0}^{\frac{4}{3}} : 12 [arctan (u)]_{0}^{\frac{4}{3}}

= 12 [arctan (u)]_{0}^{\frac{4}{3}}

Compute the boundaries:

arctan (\frac{4}{3})

= 12 arctan (\frac{4}{3})

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What is the integral from 0 to 4 of (36)/(x^2+9) ?
The integral from 0 to 4 of (36)/(x^2+9) is 12arctan(4/3)