What is the integral from 0 to 3 of 1/(4x^2+9) ?

The integral from 0 to 3 of 1/(4x^2+9) is 1/6 arctan(2)

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integral from 0 to 3 of 1/(4x^2+9)

\int_{0}^{3} \frac{1}{4 x ^{2} + 9} d x

\frac{1}{6} arctan (2)

Decimal

0.18452 \dots

Solution steps

\int_{0}^{3} \frac{1}{4 x ^{2} + 9} d x

Apply Integral Substitution

= \int_{0}^{2} \frac{1}{6 ( 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}{6} \cdot \int_{0}^{2} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{6} [arctan (u)]_{0}^{2}

Compute the boundaries:

arctan (2)

= \frac{1}{6} arctan (2)

\frac{d}{d x} (lo g_{e} (a x))

\int sec^{3} (3 x) d x

\frac{d}{d x} (1 0^{3 x})

\frac{\partial}{\partial x} (\frac{x ^{2}}{y})

x \to 2 lim (x^{2} - 3)

What is the integral from 0 to 3 of 1/(4x^2+9) ?
The integral from 0 to 3 of 1/(4x^2+9) is 1/6 arctan(2)