What is the integral of (x^2+18x-9)/(x^3-9x) ?

The integral of (x^2+18x-9)/(x^3-9x) is ln|x|-3ln|x+3|+3ln|x-3|+C

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integral of (x^2+18x-9)/(x^3-9x)

\int \frac{x ^{2} + 18 x - 9}{x ^{3} - 9 x} d x

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

Solution steps

\int \frac{x ^{2} + 18 x - 9}{x ^{3} - 9 x} d x

Take the partial fraction of

\frac{x ^{2} + 18 x - 9}{x ^{3} - 9 x} : \frac{1}{x} - \frac{3}{x + 3} + \frac{3}{x - 3}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

y = x (x + 1)^{2}, y = 0

\int_{0}^{\frac{π}{12}} cos (4 x) d x

f (x) = sin^{3} (18 x)

x \to 3 lim (x^{2} - 7)

x \to + 0 + lim (\frac{0}{x})

What is the integral of (x^2+18x-9)/(x^3-9x) ?
The integral of (x^2+18x-9)/(x^3-9x) is ln|x|-3ln|x+3|+3ln|x-3|+C