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

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

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

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int \frac{cos ( \frac{π}{x ^{41}} )}{x ^{42}} d x

\int ln (y^{2}) d y

x \to 1 - lim (\frac{2 - x}{x - 1})

x \to \infty - lim (\frac{( 8 x ^{3} + 6 x + 7 )}{( 6 x - 7 )})

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