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

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

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

\int \frac{( - 2 x - 6 )}{x ^{3} - 2 x ^{2} - 3 x} d x

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

Solution steps

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

Expand

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

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

= - 2 (- \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

- 2 (- \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 ∣) : - ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣

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

Add a constant to the solution

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

ln (49 sin (2) (x))

ln (1 + X)

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

\frac{\partial}{\partial x} (e^{x - y^{2}})

\int sin^{3} (y) d y

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