What is the integral of (3x+1)/(x^3-4x^2+x+6) ?

The integral of (3x+1)/(x^3-4x^2+x+6) is -1/6 ln|x+1|-7/3 ln|x-2|+5/2 ln|x-3|+C

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integral of (3x+1)/(x^3-4x^2+x+6)

\int \frac{3 x + 1}{x ^{3} - 4 x ^{2} + x + 6} d x

- \frac{1}{6} ln ∣ x + 1 ∣ - \frac{7}{3} ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{3 x + 1}{x ^{3} - 4 x ^{2} + x + 6} d x

Take the partial fraction of

\frac{3 x + 1}{x ^{3} - 4 x ^{2} + x + 6} : - \frac{1}{6 ( x + 1 )} - \frac{7}{3 ( x - 2 )} + \frac{5}{2 ( x - 3 )}

= \int - \frac{1}{6 ( x + 1 )} - \frac{7}{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}{6 ( x + 1 )} d x - \int \frac{7}{3 ( x - 2 )} d x + \int \frac{5}{2 ( x - 3 )} d x

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

\int \frac{7}{3 ( x - 2 )} d x = \frac{7}{3} ln ∣ x - 2 ∣

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

= - \frac{1}{6} ln ∣ x + 1 ∣ - \frac{7}{3} ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x + 1 ∣ - \frac{7}{3} ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + C

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What is the integral of (3x+1)/(x^3-4x^2+x+6) ?
The integral of (3x+1)/(x^3-4x^2+x+6) is -1/6 ln|x+1|-7/3 ln|x-2|+5/2 ln|x-3|+C