What is the integral from 0 to 7/9 of 1/(81x^2+49) ?

The integral from 0 to 7/9 of 1/(81x^2+49) is pi/(252)

integral from 0 to 7/9 of 1/(81x^2+49)

\int_{0}^{\frac{7}{9}} \frac{1}{81 x ^{2} + 49} d x

\frac{π}{252}

Decimal

0.01246 \dots

Solution steps

\int_{0}^{\frac{7}{9}} \frac{1}{81 x ^{2} + 49} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{63 ( 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}{63} \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}{63} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{63} \cdot \frac{π}{4}

Simplify

= \frac{π}{252}

\int_{x}^{5} cos (e^{t}) d t

\int_{1}^{3} \frac{1}{x - 1} d x

\int_{0}^{2} sin^{2} (x) d x

\int_{0}^{1} u d u

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

What is the integral from 0 to 7/9 of 1/(81x^2+49) ?
The integral from 0 to 7/9 of 1/(81x^2+49) is pi/(252)