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

e^{x} + e^{- x}

f (t) = sin (t)

x \to 2 lim (x^{4} + x^{2} - 4)

x \to 0 lim (\frac{x ^{3} + 8}{x + 2})

\frac{\partial}{\partial x} (2 x^{2} y + cos (x^{2}))

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