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

The integral from 0 to 2/3 of 1/(4+9x^2) is pi/(24)

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

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

\frac{π}{24}

Decimal

0.13089 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \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}^{1} \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}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{6} \cdot \frac{π}{4}

Simplify

= \frac{π}{24}

\frac{d}{d x} (\frac{e ^{x^{2}}}{cos ( 3 x )})

\frac{d}{d x} (e^{2 x} sin (7 x))

\frac{d y}{d t} + e^{t} y = - 7 e^{t}

\int sin^{2} (\frac{1}{2} π x) d x

5 ln (x^{2}), x = 1

What is the integral from 0 to 2/3 of 1/(4+9x^2) ?
The integral from 0 to 2/3 of 1/(4+9x^2) is pi/(24)