What is the integral of (x-1)/(x(x^2-5x+6)) ?

The integral of (x-1)/(x(x^2-5x+6)) is -1/6 ln|x|-1/2 ln|x-2|+2/3 ln|x-3|+C

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integral of (x-1)/(x(x^2-5x+6))

\int \frac{x - 1}{x ( x ^{2} - 5 x + 6 )} d x

- \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{2}{3} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x - 1}{x ( x ^{2} - 5 x + 6 )} d x

Take the partial fraction of

\frac{x - 1}{x ( x ^{2} - 5 x + 6 )} : - \frac{1}{6 x} - \frac{1}{2 ( x - 2 )} + \frac{2}{3 ( x - 3 )}

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

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

\int \frac{1}{2 ( x - 2 )} d x = \frac{1}{2} ln ∣ x - 2 ∣

\int \frac{2}{3 ( x - 3 )} d x = \frac{2}{3} ln ∣ x - 3 ∣

= - \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{2}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{2}{3} ln ∣ x - 3 ∣ + C

\int (3 - 5 x)^{2} d x

\frac{d y}{d x} = 2 y cos (x)

x \to 4 lim (- x^{2} - 9 x - 8)

\frac{d}{d x} (\frac{5 x}{2 x ^{- 2}})

x \to \infty lim (3 + x^{2})

What is the integral of (x-1)/(x(x^2-5x+6)) ?
The integral of (x-1)/(x(x^2-5x+6)) is -1/6 ln|x|-1/2 ln|x-2|+2/3 ln|x-3|+C