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

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

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

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

\frac{π}{4} - \frac{1}{2} arctan (\frac{1}{2})

Decimal Notation

0.55357 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{π}{4} - \frac{1}{2} arctan (\frac{1}{2})

\frac{d}{d x} ((2 x - 1)^{3} (3 x + 2)^{4})

\frac{d y}{d x} = tan (x)

n = 0 \sum \infty x^{(1 + n)}

\int_{- 1}^{0} - 5 x (3 x^{2} - 2)^{3} d x

\frac{d P}{d t} = P (1 0^{- 1} - 1 0^{- 7} P), P (0) = 7000

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