What is the integral of (x^2+36x-9)/(x^3-9x) ?

The integral of (x^2+36x-9)/(x^3-9x) is ln|x|-6ln|x+3|+6ln|x-3|+C

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integral of (x^2+36x-9)/(x^3-9x)

\int \frac{x ^{2} + 36 x - 9}{x ^{3} - 9 x} d x

ln ∣ x ∣ - 6 ln ∣ x + 3 ∣ + 6 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2} + 36 x - 9}{x ^{3} - 9 x} d x

Take the partial fraction of

\frac{x ^{2} + 36 x - 9}{x ^{3} - 9 x} : \frac{1}{x} - \frac{6}{x + 3} + \frac{6}{x - 3}

= \int \frac{1}{x} - \frac{6}{x + 3} + \frac{6}{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{6}{x + 3} d x + \int \frac{6}{x - 3} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x + 3} d x = 6 ln ∣ x + 3 ∣

\int \frac{6}{x - 3} d x = 6 ln ∣ x - 3 ∣

= ln ∣ x ∣ - 6 ln ∣ x + 3 ∣ + 6 ln ∣ x - 3 ∣

Add a constant to the solution

= ln ∣ x ∣ - 6 ln ∣ x + 3 ∣ + 6 ln ∣ x - 3 ∣ + C

\int_{0}^{\frac{π}{2}} tan^{2} (x) d x

\frac{\partial}{\partial y} (1 + x^{2})

arctan (\frac{x}{4})

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

\frac{e ^{x}}{e ^{x} + 3}

What is the integral of (x^2+36x-9)/(x^3-9x) ?
The integral of (x^2+36x-9)/(x^3-9x) is ln|x|-6ln|x+3|+6ln|x-3|+C