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

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

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

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

\frac{x ^{2}}{2} - x - 3 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + 3 ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{3}{x - 1} d x = 3 ln ∣ x - 1 ∣

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

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

Add a constant to the solution

= \frac{x ^{2}}{2} - x - 3 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + 3 ln ∣ x + 2 ∣ + C

\int \frac{11}{11 + e ^{x}} d x

\frac{d}{d t} (tan (e^{2 t}) + e^{t a n (2 t)})

x \to 0 - lim (\frac{1}{x} - \frac{1}{∣ x ∣})

\frac{d}{d x} (x^{3} - 2 x^{2} + x + 1)

\frac{\partial}{\partial x} (\frac{e ^{x}}{y + x ^{3}})

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