What is the integral of (3x-1)/(2x^2-x) ?

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

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integral of (3x-1)/(2x^2-x)

\int \frac{3 x - 1}{2 x ^{2} - x} d x

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

Solution steps

\int \frac{3 x - 1}{2 x ^{2} - x} d x

Take the partial fraction of

\frac{3 x - 1}{2 x ^{2} - x} : \frac{1}{x} + \frac{1}{2 x - 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

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