What is the integral of (x^2-6x+7)/(x^3-4x^2+x+6) ?

The integral of (x^2-6x+7)/(x^3-4x^2+x+6) is 7/6 ln|x+1|+1/3 ln|x-2|-1/2 ln|x-3|+C

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integral of (x^2-6x+7)/(x^3-4x^2+x+6)

\int \frac{x ^{2} - 6 x + 7}{x ^{3} - 4 x ^{2} + x + 6} d x

\frac{7}{6} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2} - 6 x + 7}{x ^{3} - 4 x ^{2} + x + 6} d x

Take the partial fraction of

\frac{x ^{2} - 6 x + 7}{x ^{3} - 4 x ^{2} + x + 6} : \frac{7}{6 ( x + 1 )} + \frac{1}{3 ( x - 2 )} - \frac{1}{2 ( x - 3 )}

= \int \frac{7}{6 ( x + 1 )} + \frac{1}{3 ( x - 2 )} - \frac{1}{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{7}{6 ( x + 1 )} d x + \int \frac{1}{3 ( x - 2 )} d x - \int \frac{1}{2 ( x - 3 )} d x

\int \frac{7}{6 ( x + 1 )} d x = \frac{7}{6} ln ∣ x + 1 ∣

\int \frac{1}{3 ( x - 2 )} d x = \frac{1}{3} ln ∣ x - 2 ∣

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

= \frac{7}{6} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{7}{6} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 3 ∣ + C

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What is the integral of (x^2-6x+7)/(x^3-4x^2+x+6) ?
The integral of (x^2-6x+7)/(x^3-4x^2+x+6) is 7/6 ln|x+1|+1/3 ln|x-2|-1/2 ln|x-3|+C