What is the integral of (x^2+12x-4)/(x^3-4x) ?

The integral of (x^2+12x-4)/(x^3-4x) is ln|x|-3ln|x+2|+3ln|x-2|+C

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integral of (x^2+12x-4)/(x^3-4x)

\int \frac{x ^{2} + 12 x - 4}{x ^{3} - 4 x} d x

ln ∣ x ∣ - 3 ln ∣ x + 2 ∣ + 3 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{2} + 12 x - 4}{x ^{3} - 4 x} d x

Take the partial fraction of

\frac{x ^{2} + 12 x - 4}{x ^{3} - 4 x} : \frac{1}{x} - \frac{3}{x + 2} + \frac{3}{x - 2}

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

\int \frac{3}{x - 2} d x = 3 ln ∣ x - 2 ∣

= ln ∣ x ∣ - 3 ln ∣ x + 2 ∣ + 3 ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x ∣ - 3 ln ∣ x + 2 ∣ + 3 ln ∣ x - 2 ∣ + C

\frac{\partial}{\partial x} (4 x ln (y) + y ln (z) - 4 x)

\int \frac{x ^{2}}{x - 10} d x

\frac{d y}{d x} = (4 x + y)^{2}

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

\frac{\partial}{\partial x} (ln (10 x))

What is the integral of (x^2+12x-4)/(x^3-4x) ?
The integral of (x^2+12x-4)/(x^3-4x) is ln|x|-3ln|x+2|+3ln|x-2|+C