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

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

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

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

arctan (2) + \frac{π}{2}

Decimal

2.67794 \dots

Solution steps

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

= arctan (2) - (- \frac{π}{2})

Simplify

= arctan (2) + \frac{π}{2}

\frac{\partial}{\partial y} (e^{x} + yz)

\int \frac{1}{e ^{z} + e ^{- z}} d z

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

\frac{d y}{d x}, 4 x^{2} + 9 y^{2} = 36

y^{^{'}} + 2 x y = x e^{- x^{2}}

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