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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (- x y z^{2} + 6)

\frac{d}{d x} (x sin (π) x)

x \to 0 + lim (x ln (x^{3}))

\frac{\partial}{\partial x} (\frac{2 x ^{2}}{y ^{2} + x ^{2}})

\int_{- 1}^{3} 2 x + 3 - x^{2} d x

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