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

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

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

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

arctan (x)

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

arctan (x)

= arctan (x)

\frac{\partial}{\partial x} ((x e^{x + 2 y}) (1 + 2 y))

\int \frac{- x ^{3} - x ^{2} - 4 x + 28}{x ^{3} + 4 x} d x

\int_{0}^{7} \frac{8}{9 x ^{2} + 10 x + 1} d x

\frac{\partial}{\partial x} (e^{3 y} cos (- 3 x))

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