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

The integral of 4-3(1-x^2)^{-1} is 4x-3((ln|x+1|)/2-(ln|x-1|)/2)+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 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}) + 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 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})

= 4 x - 3 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})

Add a constant to the solution

= 4 x - 3 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}) + C

\frac{d}{d x} y - y = 4

f (x) = \frac{- 2 x}{x ^{2} + 1}

\int \frac{5 - 50 x}{4 - 25 x ^{2}} d x

\int 2^{x} d x

cos (x) - sin (x)

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