What is the integral of-3(1+x^2)^{-1} ?

The integral of-3(1+x^2)^{-1} is -3arctan(x)+C

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integral of-3(1+x^2)^{-1}

\int - 3 (1 + x^{2})^{- 1} d x

- 3 arctan (x) + C

Solution steps

\int - 3 (1 + x^{2})^{- 1} d x

Take the constant out:

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

= - 3 \cdot \int (1 + x^{2})^{- 1} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= - 3 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= - 3 arctan (x)

Add a constant to the solution

= - 3 arctan (x) + C

What is the integral of-3(1+x^2)^{-1} ?
The integral of-3(1+x^2)^{-1} is -3arctan(x)+C