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

The integral of ((x^5+x^4-8))/((x^3-x)) is (x^3)/3+(x^2)/2+x+8ln|x|-4ln|x+1|-3ln|x-1|+C

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

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

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + 8 ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{5} + x ^{4} - 8}{x ^{3} - x} : x^{2} + x + 1 + \frac{8}{x} - \frac{4}{x + 1} - \frac{3}{x - 1}

= \int x^{2} + x + 1 + \frac{8}{x} - \frac{4}{x + 1} - \frac{3}{x - 1} 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 1 d x + \int \frac{8}{x} d x - \int \frac{4}{x + 1} d x - \int \frac{3}{x - 1} d x

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

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

\int 1 d x = x

\int \frac{8}{x} d x = 8 ln ∣ x ∣

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

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

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + 8 ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + 8 ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ - 3 ln ∣ x - 1 ∣ + C

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{18}} d x

\int x + 1 d x

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

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

\int - 8 x^{4} e^{x^{5}} d x

What is the integral of ((x^5+x^4-8))/((x^3-x)) ?
The integral of ((x^5+x^4-8))/((x^3-x)) is (x^3)/3+(x^2)/2+x+8ln|x|-4ln|x+1|-3ln|x-1|+C