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

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

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

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

2 x^{2} - 4 x + C

Solution steps

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

Simplify

(4 x^{2} - 2 x) x^{- 1} : \frac{4 x ^{2} - 2 x}{x}

= \int \frac{4 x ^{2} - 2 x}{x} d x

Expand

\frac{4 x ^{2} - 2 x}{x} : 4 x - \frac{2}{x}

= \int 4 x - \frac{2}{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 4 x d x - \int \frac{2}{x} d x

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

\int \frac{2}{x} d x = 4 x

= 2 x^{2} - 4 x

Add a constant to the solution

= 2 x^{2} - 4 x + C

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