What is the integral of (3x^2-5x-4)/(x^3-x) ?

The integral of (3x^2-5x-4)/(x^3-x) is 3/2 ln|x^2-1|+1/2 ln|x+1|-9/2 ln|x-1|+4ln|x|+C

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integral of (3x^2-5x-4)/(x^3-x)

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

\frac{3}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

\int \frac{5 x}{x ^{3} - x} d x = - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 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{3}{2} ln x^{2} - 1 - (- 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)) - 4 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣)

Simplify

\frac{3}{2} ln x^{2} - 1 - (- 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)) - 4 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) : \frac{3}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

= \frac{3}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

Add a constant to the solution

= \frac{3}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{9}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

\int x^{13} d x

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\int \frac{15 e ^{5 s}}{e ^{5 s} + 2} d s

What is the integral of (3x^2-5x-4)/(x^3-x) ?
The integral of (3x^2-5x-4)/(x^3-x) is 3/2 ln|x^2-1|+1/2 ln|x+1|-9/2 ln|x-1|+4ln|x|+C