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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Use the common integral:

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

= [arctan (x)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

t y^{^{'}} + 4 y = t^{2} - t + 7

t^{2} sin^{2} (t)

\frac{d y}{d x} = \frac{x}{25 y}, y (0) = - 5

2.5 (1 + e^{- 0.1 t})

\frac{d ^{2}}{d x ^{2}} (sin (3 x))

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