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

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

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

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

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

Solution steps

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

Expand

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

\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 ∣

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

Simplify

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

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

Add a constant to the solution

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

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