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

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

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integral of (x^4-4)/(x^3-x)

\int \frac{x ^{4} - 4}{x ^{3} - x} d x

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

Solution steps

\int \frac{x ^{4} - 4}{x ^{3} - x} d x

Expand

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

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

Add a constant to the solution

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

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