What is the integral of (12x^2-6x+4/x) ?

The integral of (12x^2-6x+4/x) is 4x^3-3x^2+4ln|x|+C

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integral of (12x^2-6x+4/x)

\int (12 x^{2} - 6 x + \frac{4}{x}) d x

4 x^{3} - 3 x^{2} + 4 ln ∣ x ∣ + C

Solution steps

\int 12 x^{2} - 6 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 12 x^{2} d x - \int 6 x d x + \int \frac{4}{x} d x

\int 12 x^{2} d x = 4 x^{3}

\int 6 x d x = 3 x^{2}

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= 4 x^{3} - 3 x^{2} + 4 ln ∣ x ∣

Add a constant to the solution

= 4 x^{3} - 3 x^{2} + 4 ln ∣ x ∣ + C

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What is the integral of (12x^2-6x+4/x) ?
The integral of (12x^2-6x+4/x) is 4x^3-3x^2+4ln|x|+C