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

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

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

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

= \frac{π}{2} - \frac{π}{4}

Simplify

= \frac{π}{4}

(3 - x)^{^{'}}

2 + 8 x^{\frac{3}{2}}

(t^{2} - 2)^{2}

25 y^{^{''}} - 20 y^{^{'}} + 4 y = 0

\int cos (w x) d x

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