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

The integral from 0 to 5 of 1/(25+x^2) is pi/(20)

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

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

\frac{π}{20}

Decimal Notation

0.15707 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{5 ( 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}{5} \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}{5} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{5} \cdot \frac{π}{4}

Simplify

= \frac{π}{20}

x \to 0 lim (\frac{x ^{4} + 7 x}{x})

\int (x + \frac{1}{x}) d x

\frac{df}{d x} = tan^{3} (x) - sec^{3} (x)

(- 1, - 1), (1, 0)

\frac{d}{d x} (2 x + x ln (x))

What is the integral from 0 to 5 of 1/(25+x^2) ?
The integral from 0 to 5 of 1/(25+x^2) is pi/(20)