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

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

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

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

π - \frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= π - \frac{π}{2}

\int_{- 1}^{1} x e^{2 x} d x

\int_{0}^{\infty} y \cdot e^{- y} d y

\int_{0}^{\frac{π}{4}} \frac{8}{3} sin (x) d x

\int_{6}^{\frac{1}{x}} arctan (t) d t

\int_{0}^{1} \frac{4}{3} (1 - x y) d y

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