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

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

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integral of (x^2+4x-1)/(x^3-x)

\int \frac{x ^{2} + 4 x - 1}{x ^{3} - x} d x

ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x ^{2} + 4 x - 1}{x ^{3} - x} d x

Take the partial fraction of

\frac{x ^{2} + 4 x - 1}{x ^{3} - x} : \frac{1}{x} - \frac{2}{x + 1} + \frac{2}{x - 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

f (x) = arctan (\frac{1 - x}{1 + x})

\frac{x}{7 + x}

\frac{d}{d x} (e^{(x + l n (x))})

y = sin (x), y = sin (2 x), x = 0, x = \frac{π}{3}

\frac{d}{d x} (1 - ln (cos (x)))

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