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

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

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

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

arctan (b)

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (b)

= arctan (b)

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\int_{0}^{l n (5)} e^{3 x} d x

\int_{0}^{1} (x - x^{3}) d x

\int_{1}^{e^{4}} \frac{ln ( x )}{x} d x

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