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

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

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

\int \frac{( - 9 x - 9 )}{x ( x ^{2} - 9 )} d x

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

Solution steps

\int \frac{- 9 x - 9}{x ( x ^{2} - 9 )} d x

Expand

\frac{- 9 x - 9}{x ( x ^{2} - 9 )} : - \frac{9 x}{x ( x ^{2} - 9 )} - \frac{9}{x ( x ^{2} - 9 )}

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{9 x}{x ( x ^{2} - 9 )} d x - \int \frac{9}{x ( x ^{2} - 9 )} d x

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

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

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

Simplify

- (- 3 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1)) - \frac{1}{2} (- 2 ln (x) + ln x^{2} - 9) : \frac{3}{2} ln \frac{x}{3} + 1 - \frac{3}{2} ln \frac{x}{3} - 1 + ln (x) - \frac{1}{2} ln x^{2} - 9

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

Add a constant to the solution

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

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