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

The integral from 0 to t of 1/(1+x^2) is arctan(t)

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

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

arctan (t)

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (t)

= arctan (t)

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