What is the integral of (4x^2+3x-1)/(x^3-x^2) ?

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

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integral of (4x^2+3x-1)/(x^3-x^2)

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

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

Solution steps

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

Expand

\frac{4 x ^{2} + 3 x - 1}{x ^{3} - x ^{2}} : \frac{4 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{4 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{4 x ^{2}}{x ^{3} - x ^{2}} d x = 4 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 ∣

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{y}{x + z})

\int \frac{2 x + 1}{x + 1} d x

f (x) = \frac{3}{x ^{2}}

\frac{\partial}{\partial x} (\frac{( a + c x )}{c + b})

\int cos (θ) sin (θ) d θ

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