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

The integral from 0 to 7 of 1/((49+x^2)) is pi/(28)

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

\int_{0}^{7} \frac{1}{( 49 + x ^{2} )} d x

\frac{π}{28}

Decimal Notation

0.11219 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{7 ( 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}{7} \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}{7} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{7} \cdot \frac{π}{4}

Simplify

= \frac{π}{28}

\frac{d}{d x} (sin (\frac{x}{5}))

y^{^{'''}} + 8 y^{^{''}} + 52 y^{^{'}} = 0

\int_{0}^{1} 5 e^{- 5 x} d x

\int \frac{3}{x - 1} d x

3 x^{5} - 20 x^{3}

What is the integral from 0 to 7 of 1/((49+x^2)) ?
The integral from 0 to 7 of 1/((49+x^2)) is pi/(28)