What is the integral from 60 to 90 of 1/(4+x^2) ?

The integral from 60 to 90 of 1/(4+x^2) is 1/2 (arctan(45)-arctan(30))

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integral from 60 to 90 of 1/(4+x^2)

\int_{60}^{90} \frac{1}{4 + x ^{2}} d x

\frac{1}{2} (arctan (45) - arctan (30))

Decimal

0.00555 \dots

Solution steps

\int_{60}^{90} \frac{1}{4 + x ^{2}} d x

Apply Integral Substitution

= \int_{30}^{45} \frac{1}{2 ( 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}{2} \cdot \int_{30}^{45} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{2} [arctan (u)]_{30}^{45}

Compute the boundaries:

arctan (45) - arctan (30)

= \frac{1}{2} (arctan (45) - arctan (30))

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What is the integral from 60 to 90 of 1/(4+x^2) ?
The integral from 60 to 90 of 1/(4+x^2) is 1/2 (arctan(45)-arctan(30))