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

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

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

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

π - \frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= π - \frac{π}{2}

\frac{d y}{d x} = - \frac{2 y}{x}

\frac{d}{d x} (A x e^{(- b x)})

\int \frac{1}{x ^{2} + 9 x + 20} d x

\int \frac{csc ^{2} ( x )}{1 + cot ( x )} d x

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

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