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

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

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

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

\frac{π}{8}

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{x ^{2} + 4} d x = \frac{1}{2} arctan (\frac{x}{2}) + C

Compute the boundaries:

\int_{2}^{\infty} \frac{1}{x ^{2} + 4} d x = \frac{π}{4} - \frac{π}{8}

= \frac{π}{4} - \frac{π}{8}

Simplify

= \frac{π}{8}

\frac{y ^{3}}{18} + \frac{3}{2 y}

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

\frac{4}{x}

\int (3 x^{2} + 6 x) (x^{3} + 3 x^{2} + 1)^{\frac{2}{3}} d x

\int \frac{x + 3}{x ^{2} + 4 x + 5} d x

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