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}

\int \frac{2}{( x + 2 ) ^{2} ( 2 - x )} d x

- 12 x^{2} y + y^{^{'}} = - 8 x^{2}

y^{^{''}} + 2 y^{^{'}} + y = sin (2 x)

\frac{\partial}{\partial x} (x - 2 y^{2})

\frac{d}{d x} (r (θ, x) cos (θ))

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