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

The integral of (2x^3+x-1)/(x^3+x^2-4x-4) is 2x+4/3 ln|x+1|-19/4 ln|x+2|+17/12 ln|x-2|+C

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

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

2 x + \frac{4}{3} ln ∣ x + 1 ∣ - \frac{19}{4} ln ∣ x + 2 ∣ + \frac{17}{12} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 x ^{3} + x - 1}{x ^{3} + x ^{2} - 4 x - 4} : 2 + \frac{4}{3 ( x + 1 )} - \frac{19}{4 ( x + 2 )} + \frac{17}{12 ( x - 2 )}

= \int 2 + \frac{4}{3 ( x + 1 )} - \frac{19}{4 ( x + 2 )} + \frac{17}{12 ( x - 2 )} 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 2 d x + \int \frac{4}{3 ( x + 1 )} d x - \int \frac{19}{4 ( x + 2 )} d x + \int \frac{17}{12 ( x - 2 )} d x

\int 2 d x = 2 x

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

\int \frac{19}{4 ( x + 2 )} d x = \frac{19}{4} ln ∣ x + 2 ∣

\int \frac{17}{12 ( x - 2 )} d x = \frac{17}{12} ln ∣ x - 2 ∣

= 2 x + \frac{4}{3} ln ∣ x + 1 ∣ - \frac{19}{4} ln ∣ x + 2 ∣ + \frac{17}{12} ln ∣ x - 2 ∣

Add a constant to the solution

= 2 x + \frac{4}{3} ln ∣ x + 1 ∣ - \frac{19}{4} ln ∣ x + 2 ∣ + \frac{17}{12} ln ∣ x - 2 ∣ + C

\int (16 x + 3) 8 x^{2} + 3 x d x

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

y = \frac{5 x}{1 + x ^{2}}, (4, \frac{20}{17})

\int \frac{7 - 98 x}{81 - 49 x ^{2}} d x

y = x^{0.4}

What is the integral of (2x^3+x-1)/(x^3+x^2-4x-4) ?
The integral of (2x^3+x-1)/(x^3+x^2-4x-4) is 2x+4/3 ln|x+1|-19/4 ln|x+2|+17/12 ln|x-2|+C