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

The integral of (x^2+2)/(2x^2-x) is 1/2 x+9/4 ln|2x-1|-2ln|x|-1/4+C

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

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

\frac{1}{2} x + \frac{9}{4} ln ∣ 2 x - 1 ∣ - 2 ln ∣ x ∣ - \frac{1}{4} + C

Solution steps

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

Factor

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

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

Expand

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

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

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

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

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

Simplify

\frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) + 2 (- ln ∣ x ∣ + ln ∣ 2 x - 1 ∣) : \frac{1}{2} x + \frac{9}{4} ln ∣ 2 x - 1 ∣ - 2 ln ∣ x ∣ - \frac{1}{4}

= \frac{1}{2} x + \frac{9}{4} ln ∣ 2 x - 1 ∣ - 2 ln ∣ x ∣ - \frac{1}{4}

Add a constant to the solution

= \frac{1}{2} x + \frac{9}{4} ln ∣ 2 x - 1 ∣ - 2 ln ∣ x ∣ - \frac{1}{4} + C

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What is the integral of (x^2+2)/(2x^2-x) ?
The integral of (x^2+2)/(2x^2-x) is 1/2 x+9/4 ln|2x-1|-2ln|x|-1/4+C