What is the integral of (8x-4x^{-1}) ?

The integral of (8x-4x^{-1}) is 4x^2-4ln|x|+C

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integral of (8x-4x^{-1})

\int (8 x - 4 x^{- 1}) d x

4 x^{2} - 4 ln ∣ x ∣ + C

Solution steps

\int 8 x - 4 x^{- 1} d x

4 x^{- 1} = \frac{4}{x}

= \int 8 x - \frac{4}{x} 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 8 x d x - \int \frac{4}{x} d x

\int 8 x d x = 4 x^{2}

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= 4 x^{2} - 4 ln ∣ x ∣

Add a constant to the solution

= 4 x^{2} - 4 ln ∣ x ∣ + C

ln (2 x)

\int (x + x^{2}) d x

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

\int 2 x^{- 1} + 5 d x

\int \frac{16}{25 + 9 x ^{2}} d x

What is the integral of (8x-4x^{-1}) ?
The integral of (8x-4x^{-1}) is 4x^2-4ln|x|+C