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

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

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

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

ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

= ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

\int 6 e^{x + 2 y} d y

y = \frac{4}{- 4 x + 3 + 3 x ^{2}} \frac{d y}{d x}

n = 0 \sum \infty n! (2 x - 1)^{n}

\frac{\partial}{\partial x} (4 x + 3 y)

\frac{\partial}{\partial x} (x yz - e^{z})

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