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

The integral from 0 to 2 of 1/(4+x^2) is pi/8

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

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

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \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_{0}^{1} \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)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

Simplify

= \frac{π}{8}

\frac{d}{d x} (a x)

n = 1 \sum \infty n sin (nπ)

\frac{d}{d x} ((\frac{3 x ^{2} - 2}{2 x + 3})^{3})

\frac{\partial}{\partial x} (3 x^{2} - y^{3} + x - 1)

\int \frac{( ln ( x ^{8} ) )}{x} d x

What is the integral from 0 to 2 of 1/(4+x^2) ?
The integral from 0 to 2 of 1/(4+x^2) is pi/8