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

The integral of (x^2+3)/(x^3-9x) is 1/2 ln(|x+3||x-3|)-3(1/9 ln|x|-1/18 ln|x+3|-1/18 ln|x-3|)+C

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

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

\frac{1}{2} ln (∣ x + 3 ∣ ∣ x - 3 ∣) - 3 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{18} ln ∣ x - 3 ∣) + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{x ^{3} - 9 x} d x = \frac{1}{2} ln (∣ x + 3 ∣ ∣ x - 3 ∣)

\int \frac{3}{x ^{3} - 9 x} d x = - 3 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{18} ln ∣ x - 3 ∣)

= \frac{1}{2} ln (∣ x + 3 ∣ ∣ x - 3 ∣) - 3 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{18} ln ∣ x - 3 ∣)

Add a constant to the solution

= \frac{1}{2} ln (∣ x + 3 ∣ ∣ x - 3 ∣) - 3 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{18} ln ∣ x - 3 ∣) + C

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What is the integral of (x^2+3)/(x^3-9x) ?
The integral of (x^2+3)/(x^3-9x) is 1/2 ln(|x+3||x-3|)-3(1/9 ln|x|-1/18 ln|x+3|-1/18 ln|x-3|)+C