What is the integral from 0 to 2 of (40)/(x^2+4) ?

The integral from 0 to 2 of (40)/(x^2+4) is 5pi

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integral from 0 to 2 of (40)/(x^2+4)

\int_{0}^{2} \frac{40}{x ^{2} + 4} d x

5 π

Solution steps

\int_{0}^{2} \frac{40}{x ^{2} + 4} d x

Take the constant out:

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

= 40 \cdot \int_{0}^{2} \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= 40 \cdot \int_{0}^{1} \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 40 \cdot \frac{1}{2} \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)

= 40 \cdot \frac{1}{2} [arctan (u)]_{0}^{1}

Simplify

40 \cdot \frac{1}{2} [arctan (u)]_{0}^{1} : 20 [arctan (u)]_{0}^{1}

= 20 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 20 \cdot \frac{π}{4}

Simplify

= 5 π

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What is the integral from 0 to 2 of (40)/(x^2+4) ?
The integral from 0 to 2 of (40)/(x^2+4) is 5pi