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

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

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

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

- ln ∣ x ∣ + ln ∣ x - 9 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{1}{x ^{2} - 9 x} d x

Take the partial fraction of

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

= 9 \cdot \int - \frac{1}{9 x} + \frac{1}{9 ( x - 9 )} 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

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

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{1}{9 ( x - 9 )} d x = \frac{1}{9} ln ∣ x - 9 ∣

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

Simplify

9 (- \frac{1}{9} ln ∣ x ∣ + \frac{1}{9} ln ∣ x - 9 ∣) : - ln ∣ x ∣ + ln ∣ x - 9 ∣

= - ln ∣ x ∣ + ln ∣ x - 9 ∣

Add a constant to the solution

= - ln ∣ x ∣ + ln ∣ x - 9 ∣ + C

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