What is the integral of (2x-1)/(x^3-6x^2+11x-6) ?

The integral of (2x-1)/(x^3-6x^2+11x-6) is 1/2 ln|x-1|-3ln|x-2|+5/2 ln|x-3|+C

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integral of (2x-1)/(x^3-6x^2+11x-6)

\int \frac{2 x - 1}{x ^{3} - 6 x ^{2} + 11 x - 6} d x

\frac{1}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{2 x - 1}{x ^{3} - 6 x ^{2} + 11 x - 6} d x

Take the partial fraction of

\frac{2 x - 1}{x ^{3} - 6 x ^{2} + 11 x - 6} : \frac{1}{2 ( x - 1 )} - \frac{3}{x - 2} + \frac{5}{2 ( x - 3 )}

= \int \frac{1}{2 ( x - 1 )} - \frac{3}{x - 2} + \frac{5}{2 ( 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}{2 ( x - 1 )} d x - \int \frac{3}{x - 2} d x + \int \frac{5}{2 ( x - 3 )} d x

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

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

\int \frac{5}{2 ( x - 3 )} d x = \frac{5}{2} ln ∣ x - 3 ∣

= \frac{1}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + C

x \to - \infty lim (- \frac{6}{x})

f (x) = csc^{7} (2 x)

y^{^{'}} + 2 y = e^{(4 t)}

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

\frac{d}{d x} (- 1.8 x^{2} + 14.4 x)

What is the integral of (2x-1)/(x^3-6x^2+11x-6) ?
The integral of (2x-1)/(x^3-6x^2+11x-6) is 1/2 ln|x-1|-3ln|x-2|+5/2 ln|x-3|+C