What is the integral of ((5x^3-6x+4))/(x^3-x^2) ?

The integral of ((5x^3-6x+4))/(x^3-x^2) is 5x+2ln|x|+4/x+3ln|x-1|+C

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integral of ((5x^3-6x+4))/(x^3-x^2)

\int \frac{( 5 x ^{3} - 6 x + 4 )}{x ^{3} - x ^{2}} d x

5 x + 2 ln ∣ x ∣ + \frac{4}{x} + 3 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{5 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

Take the partial fraction of

\frac{5 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} : 5 + \frac{2}{x} - \frac{4}{x ^{2}} + \frac{3}{x - 1}

= \int 5 + \frac{2}{x} - \frac{4}{x ^{2}} + \frac{3}{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 5 d x + \int \frac{2}{x} d x - \int \frac{4}{x ^{2}} d x + \int \frac{3}{x - 1} d x

\int 5 d x = 5 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

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

= 5 x + 2 ln ∣ x ∣ - (- \frac{4}{x}) + 3 ln ∣ x - 1 ∣

Simplify

= 5 x + 2 ln ∣ x ∣ + \frac{4}{x} + 3 ln ∣ x - 1 ∣

Add a constant to the solution

= 5 x + 2 ln ∣ x ∣ + \frac{4}{x} + 3 ln ∣ x - 1 ∣ + C

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What is the integral of ((5x^3-6x+4))/(x^3-x^2) ?
The integral of ((5x^3-6x+4))/(x^3-x^2) is 5x+2ln|x|+4/x+3ln|x-1|+C