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)

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

- 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

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