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

The integral of (6x)/(x^3-4x^2+x+6) is 6(-1/12 ln|x+1|-2/3 ln|x-2|+3/4 ln|x-3|)+C

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

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

6 (- \frac{1}{12} ln ∣ x + 1 ∣ - \frac{2}{3} ln ∣ x - 2 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

\frac{x}{x ^{3} - 4 x ^{2} + x + 6} : - \frac{1}{12 ( x + 1 )} - \frac{2}{3 ( x - 2 )} + \frac{3}{4 ( x - 3 )}

= 6 \cdot \int - \frac{1}{12 ( x + 1 )} - \frac{2}{3 ( x - 2 )} + \frac{3}{4 ( 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

= 6 (- \int \frac{1}{12 ( x + 1 )} d x - \int \frac{2}{3 ( x - 2 )} d x + \int \frac{3}{4 ( x - 3 )} d x)

\int \frac{1}{12 ( x + 1 )} d x = \frac{1}{12} ln ∣ x + 1 ∣

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

\int \frac{3}{4 ( x - 3 )} d x = \frac{3}{4} ln ∣ x - 3 ∣

= 6 (- \frac{1}{12} ln ∣ x + 1 ∣ - \frac{2}{3} ln ∣ x - 2 ∣ + \frac{3}{4} ln ∣ x - 3 ∣)

Add a constant to the solution

= 6 (- \frac{1}{12} ln ∣ x + 1 ∣ - \frac{2}{3} ln ∣ x - 2 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) + C

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What is the integral of (6x)/(x^3-4x^2+x+6) ?
The integral of (6x)/(x^3-4x^2+x+6) is 6(-1/12 ln|x+1|-2/3 ln|x-2|+3/4 ln|x-3|)+C