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

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

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

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

- ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= - ln ∣ x ∣ + 3 ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + C

\frac{d}{d t} (\frac{3 y ^{2} - t ^{2}}{y ^{5}})

\frac{d}{d x} (ln (\frac{4}{x x + 1}))

(e^{x^{2} - 1})^{^{'}}

\frac{\partial}{\partial x} (- 2 x y)

\frac{d}{d x} ((x^{3} - 4 x + 1) e^{x})

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