What is the integral of (-(19)/(x^2+1)) ?

The integral of (-(19)/(x^2+1)) is -19arctan(x)+C

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integral of (-(19)/(x^2+1))

\int (- \frac{19}{x ^{2} + 1}) d x

- 19 arctan (x) + C

Solution steps

\int - \frac{19}{x ^{2} + 1} d x

Take the constant out:

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

= - 19 \cdot \int \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= - 19 arctan (x)

Add a constant to the solution

= - 19 arctan (x) + C

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What is the integral of (-(19)/(x^2+1)) ?
The integral of (-(19)/(x^2+1)) is -19arctan(x)+C