What is the integral from 0 to 5 of (40)/(x^2+25) ?

The integral from 0 to 5 of (40)/(x^2+25) is 2pi

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integral from 0 to 5 of (40)/(x^2+25)

\int_{0}^{5} \frac{40}{x ^{2} + 25} d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{0}^{5} \frac{40}{x ^{2} + 25} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 40 \cdot \int_{0}^{5} \frac{1}{x ^{2} + 25} d x

Apply Integral Substitution

= 40 \cdot \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

= 40 \cdot \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)

= 40 \cdot \frac{1}{5} [arctan (u)]_{0}^{1}

Simplify

40 \cdot \frac{1}{5} [arctan (u)]_{0}^{1} : 8 [arctan (u)]_{0}^{1}

= 8 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 8 \cdot \frac{π}{4}

Simplify

= 2 π

x \to \infty lim (\frac{2}{x} - 3)

\frac{1}{1 + e ^{- x}}

\int \frac{2 x + 1}{x ^{2} + 2 x + 82} d x

x \to 2 lim (4 x^{2})

\frac{d}{d x} (e^{x^{2} - 4} + 4 x + 11)

What is the integral from 0 to 5 of (40)/(x^2+25) ?
The integral from 0 to 5 of (40)/(x^2+25) is 2pi