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

The integral of (x^3+1)/(x^3-5x^2+6x) is x+1/6 ln|x|-9/2 ln|x-2|+28/3 ln|x-3|+C

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integral of (x^3+1)/(x^3-5x^2+6x)

\int \frac{x ^{3} + 1}{x ^{3} - 5 x ^{2} + 6 x} d x

x + \frac{1}{6} ln ∣ x ∣ - \frac{9}{2} ln ∣ x - 2 ∣ + \frac{28}{3} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{3} + 1}{x ^{3} - 5 x ^{2} + 6 x} d x

Take the partial fraction of

\frac{x ^{3} + 1}{x ^{3} - 5 x ^{2} + 6 x} : 1 + \frac{1}{6 x} - \frac{9}{2 ( x - 2 )} + \frac{28}{3 ( x - 3 )}

= \int 1 + \frac{1}{6 x} - \frac{9}{2 ( x - 2 )} + \frac{28}{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 1 d x + \int \frac{1}{6 x} d x - \int \frac{9}{2 ( x - 2 )} d x + \int \frac{28}{3 ( x - 3 )} d x

\int 1 d x = x

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

\int \frac{9}{2 ( x - 2 )} d x = \frac{9}{2} ln ∣ x - 2 ∣

\int \frac{28}{3 ( x - 3 )} d x = \frac{28}{3} ln ∣ x - 3 ∣

= x + \frac{1}{6} ln ∣ x ∣ - \frac{9}{2} ln ∣ x - 2 ∣ + \frac{28}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= x + \frac{1}{6} ln ∣ x ∣ - \frac{9}{2} ln ∣ x - 2 ∣ + \frac{28}{3} ln ∣ x - 3 ∣ + C

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What is the integral of (x^3+1)/(x^3-5x^2+6x) ?
The integral of (x^3+1)/(x^3-5x^2+6x) is x+1/6 ln|x|-9/2 ln|x-2|+28/3 ln|x-3|+C