What is the integral of (6x^2-3x+1)/(x^3-x^2) ?

The integral of (6x^2-3x+1)/(x^3-x^2) is 4ln|x-1|+2ln|x|+1/x+C

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integral of (6x^2-3x+1)/(x^3-x^2)

\int \frac{6 x ^{2} - 3 x + 1}{x ^{3} - x ^{2}} d x

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

Solution steps

\int \frac{6 x ^{2} - 3 x + 1}{x ^{3} - x ^{2}} d x

Expand

\frac{6 x ^{2} - 3 x + 1}{x ^{3} - x ^{2}} : \frac{6 x ^{2}}{x ^{3} - x ^{2}} - \frac{3 x}{x ^{3} - x ^{2}} + \frac{1}{x ^{3} - x ^{2}}

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

\int \frac{6 x ^{2}}{x ^{3} - x ^{2}} d x = 6 ln ∣ x - 1 ∣

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

\int \frac{1}{x ^{3} - x ^{2}} d x = - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣

= 6 ln ∣ x - 1 ∣ - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣

Simplify

6 ln ∣ x - 1 ∣ - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣ : 4 ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x}

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

Add a constant to the solution

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

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What is the integral of (6x^2-3x+1)/(x^3-x^2) ?
The integral of (6x^2-3x+1)/(x^3-x^2) is 4ln|x-1|+2ln|x|+1/x+C