What is the integral from-1 to 0 of 1/(1+x^2) ?

The integral from-1 to 0 of 1/(1+x^2) is pi/4

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integral from-1 to 0 of 1/(1+x^2)

\int_{- 1}^{0} \frac{1}{1 + x ^{2}} d x

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{- 1}^{0} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= [arctan (x)]_{- 1}^{0}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int_{0}^{1} - 1 (x^{2} + 1) e^{- x} d x

\frac{\partial}{\partial x} (7 - 3 x^{2} - 3 y^{2})

(2.4) (6.12)

x \to 0 + lim ((x + 1)^{in x})

6 x + 5 cos (x)

What is the integral from-1 to 0 of 1/(1+x^2) ?
The integral from-1 to 0 of 1/(1+x^2) is pi/4