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

x \to 2 lim (π^{2} - 2 x + 5)

x (2 x + 7)^{\frac{2}{3}}

t \to 2 lim (8 + t^{3})

y = e^{- x} sin (x), y = - e^{- x} sin (x), 0 \leq x \leq π

\int \frac{cos ( 7 x )}{1 + sin ^{2} ( 7 x )} d x

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