What is the integral from 0 to 1 of 1/(9+16r^2) ?

The integral from 0 to 1 of 1/(9+16r^2) is 1/12 arctan(4/3)

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integral from 0 to 1 of 1/(9+16r^2)

\int_{0}^{1} \frac{1}{9 + 16 r ^{2}} d r

\frac{1}{12} arctan (\frac{4}{3})

Decimal Notation

0.07727 \dots

Solution steps

\int_{0}^{1} \frac{1}{9 + 16 r ^{2}} d r

Apply Integral Substitution

= \int_{0}^{\frac{4}{3}} \frac{1}{12 ( 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}{12} \cdot \int_{0}^{\frac{4}{3}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{12} [arctan (u)]_{0}^{\frac{4}{3}}

Compute the boundaries:

arctan (\frac{4}{3})

= \frac{1}{12} arctan (\frac{4}{3})

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What is the integral from 0 to 1 of 1/(9+16r^2) ?
The integral from 0 to 1 of 1/(9+16r^2) is 1/12 arctan(4/3)