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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{x ^{4}}{x ^{3} - x} d x = \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1)

\int \frac{x ^{3}}{x ^{3} - x} d x = \frac{x ^{3}}{x ^{2} - 1} + 2 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )})

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

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

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

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

Simplify

\frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) + \frac{x ^{3}}{x ^{2} - 1} + 2 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) - \frac{1}{2} ln x^{2} - 1 - (- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ : \frac{1}{2} x^{2} + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + \frac{x ^{3}}{x ^{2} - 1} - \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x - 1 )} - ln ∣ x ∣ - \frac{1}{2}

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

Add a constant to the solution

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

\int (\frac{32 x ^{3}}{2 x ^{4} + 5}) d x

\int 200 e^{0.1 t} + 150 e^{- 0.03 t} d t

\int \frac{32 x}{( 2 x - 1 ) ( 4 x ^{2} - 16 x + 15 )} d x

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

\int \frac{1}{3 + 9 - x ^{2}} d x

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