What is the integral of 1/((x-3)sqrt(x^2)-6x) ?

The integral of 1/((x-3)sqrt(x^2)-6x) is -1/9 ln|x|+1/9 ln|x-9|+C

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integral of 1/((x-3)sqrt(x^2)-6x)

\int \frac{1}{( x - 3 ) x ^{2} - 6 x} d x

- \frac{1}{9} ln ∣ x ∣ + \frac{1}{9} ln ∣ x - 9 ∣ + C

Solution steps

\int \frac{1}{( x - 3 ) x ^{2} - 6 x} d x

x^{2} = (x),

assuming

x \geq 0

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

Expand

(x - 3) x - 6 x : x^{2} - 9 x

= \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 )}

= \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

= - \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 ∣

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

Add a constant to the solution

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

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What is the integral of 1/((x-3)sqrt(x^2)-6x) ?
The integral of 1/((x-3)sqrt(x^2)-6x) is -1/9 ln|x|+1/9 ln|x-9|+C