What is the integral from 0 to 5/7 of 1/(49x^2+25) ?

The integral from 0 to 5/7 of 1/(49x^2+25) is pi/(140)

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integral from 0 to 5/7 of 1/(49x^2+25)

\int_{0}^{\frac{5}{7}} \frac{1}{49 x ^{2} + 25} d x

\frac{π}{140}

Decimal

0.02243 \dots

Solution steps

\int_{0}^{\frac{5}{7}} \frac{1}{49 x ^{2} + 25} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{35 ( 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}{35} \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}{35} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{35} \cdot \frac{π}{4}

Simplify

= \frac{π}{140}

\int_{1}^{e} π (ln (x))^{2} d x

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\int_{0}^{1} (2 x - 1) d x

\int_{1}^{3} \frac{18 ( ln ( x ) ) ^{5}}{x} d x

\int_{- 2}^{0} \frac{13}{6} x^{3} d x

What is the integral from 0 to 5/7 of 1/(49x^2+25) ?
The integral from 0 to 5/7 of 1/(49x^2+25) is pi/(140)