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

The integral of 1/((x^2+1)) is arctan(x)+C

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integral of 1/((x^2+1))

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

arctan (x) + C

Solution steps

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

Use the common integral:

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

= arctan (x)

Add a constant to the solution

= arctan (x) + C

\frac{\partial}{\partial y} (x^{7} + 2^{y} + x^{y})

f (x) = 2 θ - 2 sin (θ)

f (x) = \frac{6}{x}, at x = - 2

(7 x + 1)^{2}

x^{2} + 2 x, 2 x + 1

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