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

The integral of (x^2+12x+12)/(x(x^2-4)) is -3ln|x|-ln|x+2|+5ln|x-2|+C

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

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

- 3 ln ∣ x ∣ - ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 12 x + 12}{x ( x ^{2} - 4 )} : - \frac{3}{x} - \frac{1}{x + 2} + \frac{5}{x - 2}

= \int - \frac{3}{x} - \frac{1}{x + 2} + \frac{5}{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{3}{x} d x - \int \frac{1}{x + 2} d x + \int \frac{5}{x - 2} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{5}{x - 2} d x = 5 ln ∣ x - 2 ∣

= - 3 ln ∣ x ∣ - ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣

Add a constant to the solution

= - 3 ln ∣ x ∣ - ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣ + C

\frac{d}{d x} (\frac{8 x ^{2} + 6 x + 6}{x})

\int 16 sin^{2} (x co) s^{2} x d x

y^{^{''}} + 4 y = sec^{2} (2 x)

\int 5 tan^{3} (θ) d θ

\frac{1}{s ^{2} + s + 1}

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