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

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

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

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

- arctan (y) + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= - arctan (y)

Add a constant to the solution

= - arctan (y) + C

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