What is the integral from 0 to 4 of 1/(36x^2+25) ?

The integral from 0 to 4 of 1/(36x^2+25) is 1/30 arctan(24/5)

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integral from 0 to 4 of 1/(36x^2+25)

\int_{0}^{4} \frac{1}{36 x ^{2} + 25} d x

\frac{1}{30} arctan (\frac{24}{5})

Decimal

0.04551 \dots

Solution steps

\int_{0}^{4} \frac{1}{36 x ^{2} + 25} d x

Apply Integral Substitution

= \int_{0}^{\frac{24}{5}} \frac{1}{30 ( 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}{30} \cdot \int_{0}^{\frac{24}{5}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{30} [arctan (u)]_{0}^{\frac{24}{5}}

Compute the boundaries:

arctan (\frac{24}{5})

= \frac{1}{30} arctan (\frac{24}{5})

What is the integral from 0 to 4 of 1/(36x^2+25) ?
The integral from 0 to 4 of 1/(36x^2+25) is 1/30 arctan(24/5)