What is the integral of (4x^2-14x-6)/(x^3-2x^2-3x) ?

The integral of (4x^2-14x-6)/(x^3-2x^2-3x) is 3ln|x+1|-ln|x-3|+2ln|x|+C

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integral of (4x^2-14x-6)/(x^3-2x^2-3x)

\int \frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

3 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

Expand

\frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} : \frac{4 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} - \frac{14 x}{x ^{3} - 2 x ^{2} - 3 x} - \frac{6}{x ^{3} - 2 x ^{2} - 3 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{4 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} d x - \int \frac{14 x}{x ^{3} - 2 x ^{2} - 3 x} d x - \int \frac{6}{x ^{3} - 2 x ^{2} - 3 x} d x

\int \frac{4 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} d x = ln ∣ x + 1 ∣ + 3 ln ∣ x - 3 ∣

\int \frac{14 x}{x ^{3} - 2 x ^{2} - 3 x} d x = 14 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

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

= ln ∣ x + 1 ∣ + 3 ln ∣ x - 3 ∣ - 14 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) - 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣)

Simplify

ln ∣ x + 1 ∣ + 3 ln ∣ x - 3 ∣ - 14 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) - 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣) : 3 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣

= 3 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣ + C

n = 1 \sum \infty 1 - cos (\frac{1}{n})

x \to + 1 + lim (\frac{x - 1}{x - 1})

\frac{( y ^{^{'}} - e ^{- t} + 6 )}{y} = - 6

- 0.1 x^{3} + 600 x - 20

x \to 0 lim (cos (x + sin (x)))

What is the integral of (4x^2-14x-6)/(x^3-2x^2-3x) ?
The integral of (4x^2-14x-6)/(x^3-2x^2-3x) is 3ln|x+1|-ln|x-3|+2ln|x|+C