What is the integral of (x^2+24x-9)/(x^3-9x) ?

The integral of (x^2+24x-9)/(x^3-9x) is ln|x|-4ln|x+3|+4ln|x-3|+C

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integral of (x^2+24x-9)/(x^3-9x)

\int \frac{x ^{2} + 24 x - 9}{x ^{3} - 9 x} d x

ln ∣ x ∣ - 4 ln ∣ x + 3 ∣ + 4 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2} + 24 x - 9}{x ^{3} - 9 x} d x

Take the partial fraction of

\frac{x ^{2} + 24 x - 9}{x ^{3} - 9 x} : \frac{1}{x} - \frac{4}{x + 3} + \frac{4}{x - 3}

= \int \frac{1}{x} - \frac{4}{x + 3} + \frac{4}{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{4}{x + 3} d x + \int \frac{4}{x - 3} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{4}{x + 3} d x = 4 ln ∣ x + 3 ∣

\int \frac{4}{x - 3} d x = 4 ln ∣ x - 3 ∣

= ln ∣ x ∣ - 4 ln ∣ x + 3 ∣ + 4 ln ∣ x - 3 ∣

Add a constant to the solution

= ln ∣ x ∣ - 4 ln ∣ x + 3 ∣ + 4 ln ∣ x - 3 ∣ + C

x \to \frac{π}{2} lim (- cos (x))

y = \frac{6 x ^{2} + 6 x + 2}{x}

\frac{d}{d x} (\frac{3 x ^{2}}{5 x ^{2} + 7 x})

5 cos (x)

\frac{\partial}{\partial x} (x ln (t))

What is the integral of (x^2+24x-9)/(x^3-9x) ?
The integral of (x^2+24x-9)/(x^3-9x) is ln|x|-4ln|x+3|+4ln|x-3|+C