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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{\infty} \frac{1}{x ^{2} + 1} d x = \frac{π}{2} - 0

= \frac{π}{2} - 0

Simplify

= \frac{π}{2}

\frac{d y}{d x} = 29 - y

(1 + 4 x)^{\frac{3}{2}}, at x = 2

\int \frac{e ^{x}}{25 - e ^{2 x}} d x

x \to + (- 1) lim (\frac{k x + 3}{4})

\frac{d}{d x} (a x^{2} e^{5 x})

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