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

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

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

integral

\frac{- 1}{( 1 + x ^{2} )}

- arctan (x) + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= - arctan (x)

Add a constant to the solution

= - arctan (x) + C

\frac{( 3 x ^{2} - 1 )}{( x ^{2} + 1 )}

x^{2} x + 2

\frac{1 + 1}{x ^{4}}

\frac{e ^{x} + 1}{e ^{x}}

e^{7} y

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