What is the integral of (15x^2-4x-28)/(x^3-4x) ?

The integral of (15x^2-4x-28)/(x^3-4x) is

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integral of (15x^2-4x-28)/(x^3-4x)

\int \frac{15 x ^{2} - 4 x - 28}{x ^{3} - 4 x} d x

\frac{15}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1 + 7 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 2 ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{15 x ^{2} - 4 x - 28}{x ^{3} - 4 x} d x

Expand

\frac{15 x ^{2} - 4 x - 28}{x ^{3} - 4 x} : \frac{15 x ^{2}}{x ^{3} - 4 x} - \frac{4 x}{x ^{3} - 4 x} - \frac{28}{x ^{3} - 4 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{15 x ^{2}}{x ^{3} - 4 x} d x - \int \frac{4 x}{x ^{3} - 4 x} d x - \int \frac{28}{x ^{3} - 4 x} d x

\int \frac{15 x ^{2}}{x ^{3} - 4 x} d x = \frac{15}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣)

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

\int \frac{28}{x ^{3} - 4 x} d x = - 28 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣)

= \frac{15}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - (- ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1) - (- 28 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣))

Simplify

\frac{15}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) - (- ln \frac{x}{2} + 1 + ln \frac{x}{2} - 1) - (- 28 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln ∣ x + 2 ∣ - \frac{1}{8} ln ∣ x - 2 ∣)) : \frac{15}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1 + 7 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 2 ∣ - \frac{7}{2} ln ∣ x - 2 ∣

= \frac{15}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1 + 7 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 2 ∣ - \frac{7}{2} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{15}{2} ln (∣ x + 2 ∣ ∣ x - 2 ∣) + ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1 + 7 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 2 ∣ - \frac{7}{2} ln ∣ x - 2 ∣ + C

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What is the integral of (15x^2-4x-28)/(x^3-4x) ?
The integral of (15x^2-4x-28)/(x^3-4x) is