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{\partial}{\partial x} (3 y^{5} x^{2} + 8 y x^{3})

x \to 3 - lim (\frac{7 - x - 4}{x - 3})

(1 + x^{14}) (\frac{d y}{d x}) + 14 x^{13} y = 11 x^{5}

3^{x^{2} - 1}, at x = 2

\int 3^{5 x - 2} d x

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)