What is the integral from 2 to 2sqrt(3 of) 1/(4+z^2) ?

The integral from 2 to 2sqrt(3 of) 1/(4+z^2) is pi/(24)

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integral from 2 to 2sqrt(3 of) 1/(4+z^2)

\int_{2}^{23} \frac{1}{4 + z ^{2}} d z

\frac{π}{24}

Decimal

0.13089 \dots

Solution steps

\int_{2}^{23} \frac{1}{4 + z ^{2}} d z

Apply Integral Substitution

= \int_{1}^{3} \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_{1}^{3} \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)]_{1}^{3}

Compute the boundaries:

\frac{π}{12}

= \frac{1}{2} \cdot \frac{π}{12}

Simplify

= \frac{π}{24}

\int_{0}^{5} x^{2} - 4 d x

\int_{0}^{1} \frac{1}{x ^{\frac{2}{3}}} d x

\int_{0}^{1} x 3^{x^{2}} d x

\int_{- 2}^{0} \frac{1}{1 - x} d x

\int_{- π}^{π} e^{- x} d x

What is the integral from 2 to 2sqrt(3 of) 1/(4+z^2) ?
The integral from 2 to 2sqrt(3 of) 1/(4+z^2) is pi/(24)