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

\int \frac{6 ( 1 + x )}{x ( x ^{2} + 7 x + 6 )} d x

\frac{\partial}{\partial x} (cos^{4} (x^{3} y^{5}))

x \to - 8 lim (x^{2} - 3)

f (x) = 4 x^{3} - 8 x

a e^{b t}

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