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

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

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

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

- \frac{1}{6} ln ∣ x ∣ + \frac{1}{6} ln ∣ x - 6 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} - 6 x} : - \frac{1}{6 x} + \frac{1}{6 ( x - 6 )}

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

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

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

= - \frac{1}{6} ln ∣ x ∣ + \frac{1}{6} ln ∣ x - 6 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x ∣ + \frac{1}{6} ln ∣ x - 6 ∣ + C

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\int sec (t) (2 sec (t) + 7 tan (t)) d t

\frac{d y}{d x} = x + 4 y - 2

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\frac{\partial}{\partial x} (x^{3} - x sin (y))

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