What is the integral of-1/2 ln|x|-8x+2x^2+C^1 ?

The integral of-1/2 ln|x|-8x+2x^2+C^1 is -1/2 (xln|x|-x)-4x^2+(2x^3)/3+Cx+C

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integral of-1/2 ln|x|-8x+2x^2+C^1

\int - \frac{1}{2} ln ∣ x ∣ - 8 x + 2 x^{2} + C^{1} d x

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

Solution steps

\int - \frac{1}{2} ln ∣ x ∣ - 8 x + 2 x^{2} + C^{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 \frac{1}{2} ln ∣ x ∣ d x - \int 8 x d x + \int 2 x^{2} d x + \int C^{1} d x

\int \frac{1}{2} ln ∣ x ∣ d x = \frac{1}{2} (x ln ∣ x ∣ - x)

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

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

\int C^{1} d x = C x

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

Add a constant to the solution

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

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\frac{d y}{d x} = y ln (x)

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

\int (1 - x) (e^{- x} + 1) d x

What is the integral of-1/2 ln|x|-8x+2x^2+C^1 ?
The integral of-1/2 ln|x|-8x+2x^2+C^1 is -1/2 (xln|x|-x)-4x^2+(2x^3)/3+Cx+C