What is the integral of (5x^2-12x-12)/(x^3-4x) ?

The integral of (5x^2-12x-12)/(x^3-4x) is

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integral of (5x^2-12x-12)/(x^3-4x)

\int \frac{5 x ^{2} - 12 x - 12}{x ^{3} - 4 x} d x

\frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + 6 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{5 x ^{2} - 12 x - 12}{x ^{3} - 4 x} d x

Expand

\frac{5 x ^{2} - 12 x - 12}{x ^{3} - 4 x} : \frac{5 x ^{2}}{x ^{3} - 4 x} - \frac{12 x}{x ^{3} - 4 x} - \frac{12}{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{5 x ^{2}}{x ^{3} - 4 x} d x - \int \frac{12 x}{x ^{3} - 4 x} d x - \int \frac{12}{x ^{3} - 4 x} d x

\int \frac{5 x ^{2}}{x ^{3} - 4 x} d x = \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣)

\int \frac{12 x}{x ^{3} - 4 x} d x = - 6 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)

\int \frac{12}{x ^{3} - 4 x} d x = - 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣)

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - (- 6 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1)) - (- 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣))

Simplify

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + 6 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣)

Add a constant to the solution

= \frac{5}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + 6 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣) + C

\int \frac{2 x ^{5} - x ^{3} - 1}{x ^{3} - 4 x} d x

x \to 1 + lim (\frac{2 x}{x - 1})

\int_{0}^{2} (2 x^{2} + 1) d x

\int e^{x + 2} d x

\frac{\partial}{\partial m} (m x^{b})

What is the integral of (5x^2-12x-12)/(x^3-4x) ?
The integral of (5x^2-12x-12)/(x^3-4x) is