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

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

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

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

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

Solution steps

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

Expand

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

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

Add a constant to the solution

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

\int \frac{5 - x e ^{x}}{x} d x

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

\int (5 x^{3} - 9 x^{2} + 7 x + 3) e^{- x} d x

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