What is the integral of (6x^3-9x+7)/(x^3-x^2) ?

The integral of (6x^3-9x+7)/(x^3-x^2) is 6x+2ln|x|+7/x+4ln|x-1|+C

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integral of (6x^3-9x+7)/(x^3-x^2)

\int \frac{6 x ^{3} - 9 x + 7}{x ^{3} - x ^{2}} d x

6 x + 2 ln ∣ x ∣ + \frac{7}{x} + 4 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{6 x ^{3} - 9 x + 7}{x ^{3} - x ^{2}} d x

Take the partial fraction of

\frac{6 x ^{3} - 9 x + 7}{x ^{3} - x ^{2}} : 6 + \frac{2}{x} - \frac{7}{x ^{2}} + \frac{4}{x - 1}

= \int 6 + \frac{2}{x} - \frac{7}{x ^{2}} + \frac{4}{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 6 d x + \int \frac{2}{x} d x - \int \frac{7}{x ^{2}} d x + \int \frac{4}{x - 1} d x

\int 6 d x = 6 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{7}{x ^{2}} d x = - \frac{7}{x}

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= 6 x + 2 ln ∣ x ∣ - (- \frac{7}{x}) + 4 ln ∣ x - 1 ∣

Simplify

= 6 x + 2 ln ∣ x ∣ + \frac{7}{x} + 4 ln ∣ x - 1 ∣

Add a constant to the solution

= 6 x + 2 ln ∣ x ∣ + \frac{7}{x} + 4 ln ∣ x - 1 ∣ + C

What is the integral of (6x^3-9x+7)/(x^3-x^2) ?
The integral of (6x^3-9x+7)/(x^3-x^2) is 6x+2ln|x|+7/x+4ln|x-1|+C