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

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

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

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

arctan (R)

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (R)

= arctan (R)

\int sin (2 x) - cos (2 x) d x

\int \frac{( x - b )}{x} d x

\frac{\partial}{\partial x} (sin (x) e^{y x^{2}} x)

\int \frac{z}{1 + z ^{2}} d z

\int e^{x^{2} - 2 x} d x

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