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

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

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

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

arctan (2)

Solution steps

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

Use the common integral:

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

= [arctan (x)]_{0}^{2}

Compute the boundaries:

arctan (2)

= arctan (2)

\frac{d}{d x} (\frac{e ^{x}}{2 x})

\frac{d y}{d x} = 500 e^{\frac{1}{10} l n (\frac{625}{500}) 50}

\int e^{6 x} 2^{9 x} d x

\frac{d}{d t} ((cos (t))^{2})

\frac{\partial}{\partial y} (x y e^{- 4 y})

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