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

The integral from 0 to 2 of 1/(8+2x^2) is pi/(16)

Popular >

integral from 0 to 2 of 1/(8+2x^2)

\int_{0}^{2} \frac{1}{8 + 2 x ^{2}} d x

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

\int_{0}^{2} \frac{1}{8 + 2 x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{4 ( 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}{4} \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}{4} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Simplify

= \frac{π}{16}

3 ln (x), x ln (x), 1, 2

\frac{d y}{d x} = - \frac{e ^{x}}{8}

\int sec^{2} (3 x) tan^{3} (3 x) d x

x \cdot 6^{x}

\frac{\partial}{\partial y} (\frac{x}{7 + y})

What is the integral from 0 to 2 of 1/(8+2x^2) ?
The integral from 0 to 2 of 1/(8+2x^2) is pi/(16)