What is the integral of (9x^3-3x+1)/(x^3-x^2) ?

The integral of (9x^3-3x+1)/(x^3-x^2) is (9x^2)/(x-1)+7ln|x-1|-9/(x-1)+2ln|x|+1/x+C

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integral of (9x^3-3x+1)/(x^3-x^2)

\int \frac{9 x ^{3} - 3 x + 1}{x ^{3} - x ^{2}} d x

\frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

\int \frac{9 x ^{3} - 3 x + 1}{x ^{3} - x ^{2}} d x

Expand

\frac{9 x ^{3} - 3 x + 1}{x ^{3} - x ^{2}} : \frac{9 x ^{3}}{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{9 x ^{3}}{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{9 x ^{3}}{x ^{3} - x ^{2}} d x = 9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{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 ∣

= 9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣

Simplify

9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 3 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣ : \frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{1}{x}

= \frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{1}{x}

Add a constant to the solution

= \frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{1}{x} + C

What is the integral of (9x^3-3x+1)/(x^3-x^2) ?
The integral of (9x^3-3x+1)/(x^3-x^2) is (9x^2)/(x-1)+7ln|x-1|-9/(x-1)+2ln|x|+1/x+C