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

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

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

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

arctan (x) + \frac{π}{4}

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (x) + \frac{π}{4}

= arctan (x) + \frac{π}{4}

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