What is the integral of (3x^3-x-2)/(x^4-16) ?

The integral of (3x^3-x-2)/(x^4-16) is

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integral of (3x^3-x-2)/(x^4-16)

\int \frac{3 x ^{3} - x - 2}{x ^{4} - 16} d x

\frac{3}{4} ln x^{4} - 16 + \frac{1}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) - 2 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{3 x ^{3} - x - 2}{x ^{4} - 16} d x

Expand

\frac{3 x ^{3} - x - 2}{x ^{4} - 16} : \frac{3 x ^{3}}{x ^{4} - 16} - \frac{x}{x ^{4} - 16} - \frac{2}{x ^{4} - 16}

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 ^{3}}{x ^{4} - 16} d x - \int \frac{x}{x ^{4} - 16} d x - \int \frac{2}{x ^{4} - 16} d x

\int \frac{3 x ^{3}}{x ^{4} - 16} d x = \frac{3}{4} ln x^{4} - 16

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

\int \frac{2}{x ^{4} - 16} d x = 2 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣)

= \frac{3}{4} ln x^{4} - 16 - (- \frac{1}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1)) - 2 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣)

Simplify

= \frac{3}{4} ln x^{4} - 16 + \frac{1}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) - 2 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣)

Add a constant to the solution

= \frac{3}{4} ln x^{4} - 16 + \frac{1}{16} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) - 2 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣) + C

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What is the integral of (3x^3-x-2)/(x^4-16) ?
The integral of (3x^3-x-2)/(x^4-16) is