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

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

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

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

arctan (A) - arctan (- A)

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (A) - arctan (- A)

= arctan (A) - arctan (- A)

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