What is the integral of (9x^3-6x+4)/(x^3-x^2) ?

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

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integral of (9x^3-6x+4)/(x^3-x^2)

\int \frac{9 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

\frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x} + C

Solution steps

\int \frac{9 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

Expand

\frac{9 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} : \frac{9 x ^{3}}{x ^{3} - x ^{2}} - \frac{6 x}{x ^{3} - x ^{2}} + \frac{4}{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{6 x}{x ^{3} - x ^{2}} d x + \int \frac{4}{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{6 x}{x ^{3} - x ^{2}} d x = 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

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

= 9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 4 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

Simplify

9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 4 (- 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{4}{x}

= \frac{9 x ^{2}}{x - 1} + 7 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{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{4}{x} + C

\frac{\partial}{\partial t} (e^{- 2 t} cos (2 x))

\int (2 e^{3 x}) d x

\int (4 x^{2} + \frac{1}{( 4 x ) ^{2}}) d x

y^{^{'}} = 9 x 7 y + 8, y (0) = 3

\frac{d}{d x} (\frac{x \cdot cos ( x ) + π}{π - x})

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