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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

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

Simplify

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

Add a constant to the solution

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

\int_{0}^{π} sin^{2} (x) d x

\frac{\partial}{\partial x} (ln (4 - x^{2}))

y = 4 x^{2} - x^{3}, (1, 3)

(0.1) (1.1)

x \to a lim (0)

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