What is the integral of (x^2+12x-9)/(x^3-9x) ?

The integral of (x^2+12x-9)/(x^3-9x) is ln|x|-2ln|x+3|+2ln|x-3|+C

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integral of (x^2+12x-9)/(x^3-9x)

\int \frac{x ^{2} + 12 x - 9}{x ^{3} - 9 x} d x

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

Solution steps

\int \frac{x ^{2} + 12 x - 9}{x ^{3} - 9 x} d x

Take the partial fraction of

\frac{x ^{2} + 12 x - 9}{x ^{3} - 9 x} : \frac{1}{x} - \frac{2}{x + 3} + \frac{2}{x - 3}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

y^{^{''}} + 13 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int - 6 sin^{2} (t) d t

\frac{d}{d x} (x^{2} (x - 1))

x \to \infty lim (\frac{1}{1})

n = 0 \sum \infty \frac{n + 1}{2 n - 1}

What is the integral of (x^2+12x-9)/(x^3-9x) ?
The integral of (x^2+12x-9)/(x^3-9x) is ln|x|-2ln|x+3|+2ln|x-3|+C