What is the integral of (x^5+x^4-8)/(x^3-4x) ?

The integral of (x^5+x^4-8)/(x^3-4x) is (x^3)/3+(x^2)/2+4x+2ln|x|-3ln|x+2|+5ln|x-2|+C

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integral of (x^5+x^4-8)/(x^3-4x)

\int \frac{x ^{5} + x ^{4} - 8}{x ^{3} - 4 x} d x

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 4 x + 2 ln ∣ x ∣ - 3 ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{5} + x ^{4} - 8}{x ^{3} - 4 x} d x

Take the partial fraction of

\frac{x ^{5} + x ^{4} - 8}{x ^{3} - 4 x} : x^{2} + x + 4 + \frac{2}{x} - \frac{3}{x + 2} + \frac{5}{x - 2}

= \int x^{2} + x + 4 + \frac{2}{x} - \frac{3}{x + 2} + \frac{5}{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^{2} d x + \int x d x + \int 4 d x + \int \frac{2}{x} d x - \int \frac{3}{x + 2} d x + \int \frac{5}{x - 2} d x

\int x^{2} d x = \frac{x ^{3}}{3}

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

\int 4 d x = 4 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

\int \frac{5}{x - 2} d x = 5 ln ∣ x - 2 ∣

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 4 x + 2 ln ∣ x ∣ - 3 ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + 4 x + 2 ln ∣ x ∣ - 3 ln ∣ x + 2 ∣ + 5 ln ∣ x - 2 ∣ + C

\int x^{2} (x + 1)^{2} d x

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

5 x^{2} y

x \to 0 lim (\frac{7 x ^{3} - x ^{2}}{x})

(x^{2} + y^{2}) d x - (2 \cdot x \cdot y) d y = 0

What is the integral of (x^5+x^4-8)/(x^3-4x) ?
The integral of (x^5+x^4-8)/(x^3-4x) is (x^3)/3+(x^2)/2+4x+2ln|x|-3ln|x+2|+5ln|x-2|+C