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

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

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

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

- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

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

= - \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣ + C

x^{2} - \frac{2}{x ^{3}}

\frac{x ^{3}}{3} - 11 x, 5 x

\frac{d}{d x} ((2 x + 4)^{2})

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

\int \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}} d x

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