What is the integral from 1 to infinity of 1/(x^2+5) ?

The integral from 1 to infinity of 1/(x^2+5) is (pi-2arctan((sqrt(5))/5))/(2sqrt(5))

Popular >

integral from 1 to infinity of 1/(x^2+5)

\int_{1}^{\infty} \frac{1}{x ^{2} + 5} d x

\frac{π - 2 arctan ( \frac{5}{5} )}{2 5}

Decimal Notation

0.51441 \dots

Solution steps

\int_{1}^{\infty} \frac{1}{x ^{2} + 5} d x

Compute the indefinite integral:

\int \frac{1}{x ^{2} + 5} d x = \frac{1}{5} arctan (\frac{x}{5}) + C

Compute the boundaries:

\int_{1}^{\infty} \frac{1}{x ^{2} + 5} d x = \frac{π}{2 5} - \frac{1}{5} arctan (\frac{1}{5})

= \frac{π}{2 5} - \frac{1}{5} arctan (\frac{1}{5})

Simplify

= \frac{π - 2 arctan ( \frac{5}{5} )}{2 5}

5^{t}

f (x) = x + 3, at x = 1

\frac{d}{d x} (\frac{ln ( x ^{5} )}{ln ( x ^{4} )})

\int_{- 2}^{2} (8 - 2 y^{2}) d y

x \to 0 lim (x^{2} + x - x)

What is the integral from 1 to infinity of 1/(x^2+5) ?
The integral from 1 to infinity of 1/(x^2+5) is (pi-2arctan((sqrt(5))/5))/(2sqrt(5))