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

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

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integral from 0 to infinity of 8/(x^2+4)

\int_{0}^{\infty} \frac{8}{x ^{2} + 4} d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{0}^{\infty} \frac{8}{x ^{2} + 4} d x

Compute the indefinite integral:

\int \frac{8}{x ^{2} + 4} d x = 4 arctan (\frac{x}{2}) + C

Compute the boundaries:

\int_{0}^{\infty} \frac{8}{x ^{2} + 4} d x = 2 π - 0

= 2 π - 0

Simplify

= 2 π

\frac{d}{d x} (x \cdot x^{2} + 1)

y = x x, (1, 1)

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

\frac{\partial}{\partial x} (x^{\frac{1}{4}})

x \to 1 lim (2 x + 3)

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