What is the integral from 0 to 6 of 1/(36+x^2) ?

The integral from 0 to 6 of 1/(36+x^2) is pi/(24)

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integral from 0 to 6 of 1/(36+x^2)

\int_{0}^{6} \frac{1}{36 + x ^{2}} d x

\frac{π}{24}

Decimal Notation

0.13089 \dots

Solution steps

\int_{0}^{6} \frac{1}{36 + x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{6 ( 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}{6} \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}{6} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{6} \cdot \frac{π}{4}

Simplify

= \frac{π}{24}

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\frac{4}{s ^{2}} - \frac{10}{s}

\frac{d}{d x} (\frac{1}{1 + e ^{- x}})

h \to 0 - lim (\frac{2 ^{h} - 1}{h})

What is the integral from 0 to 6 of 1/(36+x^2) ?
The integral from 0 to 6 of 1/(36+x^2) is pi/(24)