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

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

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

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

\frac{π}{2}

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

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