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

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

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

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

- arctan (u) + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= - arctan (u)

Add a constant to the solution

= - arctan (u) + C