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

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

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

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

\frac{π}{4}

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int \frac{2 x + 7}{x ^{2} + 2 x - 4} d x

\frac{d}{d x} (\frac{cos ( x )}{1 + co tan ( x )})

\int 7 e^{x} + 2 sec^{2} (x) d x

\frac{d}{d x} (x (a + b x) e^{3 x})

\frac{d}{d x} (e^{- \frac{1}{x ^{3}}})

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