What is the integral from 0 to 2sqrt(3 of) 1/(x^2+4) ?

The integral from 0 to 2sqrt(3 of) 1/(x^2+4) is pi/6

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integral from 0 to 2sqrt(3 of) 1/(x^2+4)

\int_{0}^{23} \frac{1}{x ^{2} + 4} d x

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{23} \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= \int_{0}^{3} \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

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

Use the common integral:

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

= \frac{1}{2} [arctan (u)]_{0}^{3}

Compute the boundaries:

\frac{π}{3}

= \frac{1}{2} \cdot \frac{π}{3}

Simplify

= \frac{π}{6}

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What is the integral from 0 to 2sqrt(3 of) 1/(x^2+4) ?
The integral from 0 to 2sqrt(3 of) 1/(x^2+4) is pi/6