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

The integral of ((2x^2-2x+2)/(2x)) is 1/2 x^2-x+ln|x|+C

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integral of ((2x^2-2x+2)/(2x))

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

\frac{1}{2} x^{2} - x + ln ∣ x ∣ + C

Solution steps

\int \frac{2 x ^{2} - 2 x + 2}{2 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{2 x ^{2} - 2 x + 2}{x} d x

Expand

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

= \frac{1}{2} \cdot \int 2 x - 2 + \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

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

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

\int 2 d x = 2 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

Simplify

\frac{1}{2} (x^{2} - 2 x + 2 ln ∣ x ∣) : \frac{1}{2} x^{2} - x + ln ∣ x ∣

= \frac{1}{2} x^{2} - x + ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2} x^{2} - x + ln ∣ x ∣ + C

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