What is the integral of (4-3(1+x^2)^{-1}) ?

The integral of (4-3(1+x^2)^{-1}) is 4x-3arctan(x)+C

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integral of (4-3(1+x^2)^{-1})

\int (4 - 3 (1 + x^{2})^{- 1}) d x

4 x - 3 arctan (x) + C

Solution steps

\int 4 - 3 (1 + x^{2})^{- 1} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 d x - \int 3 (1 + x^{2})^{- 1} d x

\int 4 d x = 4 x

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

= 4 x - 3 arctan (x)

Add a constant to the solution

= 4 x - 3 arctan (x) + C

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What is the integral of (4-3(1+x^2)^{-1}) ?
The integral of (4-3(1+x^2)^{-1}) is 4x-3arctan(x)+C