What is the integral from 0 to 4 of 1/(4t^2+9) ?

The integral from 0 to 4 of 1/(4t^2+9) is 1/6 arctan(8/3)

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integral from 0 to 4 of 1/(4t^2+9)

\int_{0}^{4} \frac{1}{4 t ^{2} + 9} d t

\frac{1}{6} arctan (\frac{8}{3})

Decimal

0.20200 \dots

Solution steps

\int_{0}^{4} \frac{1}{4 t ^{2} + 9} d t

Apply Integral Substitution

= \int_{0}^{\frac{8}{3}} \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}^{\frac{8}{3}} \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}^{\frac{8}{3}}

Compute the boundaries:

arctan (\frac{8}{3})

= \frac{1}{6} arctan (\frac{8}{3})

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What is the integral from 0 to 4 of 1/(4t^2+9) ?
The integral from 0 to 4 of 1/(4t^2+9) is 1/6 arctan(8/3)