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

The integral of (x^2+12x+12)/(x(x+2)(x-2)) is

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

\int \frac{x ^{2} + 12 x + 12}{x ( x + 2 ) ( x - 2 )} d x

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

Solution steps

\int \frac{x ^{2} + 12 x + 12}{x ( x + 2 ) ( x - 2 )} d x

Expand

\frac{x ^{2} + 12 x + 12}{x ( x + 2 ) ( x - 2 )} : \frac{x ^{2}}{x ( x + 2 ) ( x - 2 )} + \frac{12 x}{x ( x + 2 ) ( x - 2 )} + \frac{12}{x ( x + 2 ) ( x - 2 )}

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 ( x + 2 ) ( x - 2 )} d x + \int \frac{12 x}{x ( x + 2 ) ( x - 2 )} d x + \int \frac{12}{x ( x + 2 ) ( x - 2 )} d x

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

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

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

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

Add a constant to the solution

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

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

\int \frac{e ^{x}}{1 - e ^{2 x}} d x

\int x - 1 d x

\int_{1}^{2} x x - 1 d x

\frac{1}{s ^{2} ( s + \frac{1}{15} )}

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