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

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

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

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

\frac{1}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1 + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣) + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{x ^{3} - 4 x} d x = \frac{1}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣)

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

\int \frac{2}{x ^{3} - 4 x} d x = - 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣)

= \frac{1}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1 - (- 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣))

Simplify

= \frac{1}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1 + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣)

Add a constant to the solution

= \frac{1}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1 + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣) + C

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