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

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

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

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

arctan (x) - \frac{π}{4}

Solution steps

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

Use the common integral:

\int \frac{1}{t ^{2} + 1} 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/(t^2+1) ?
The integral from 1 to x of 1/(t^2+1) is arctan(x)-pi/4