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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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

x (x - 14)

x \to \infty lim (\frac{x}{16})

\int \frac{cos ( a x )}{b + sin ( a x )} d x

y^{^{''}} + 2 x y^{^{'}} = 0

\int sec^{2} (x t a) n^{3} x d x

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