What is the integral of (9x^3-3x^2+20x+20)/(x^4-16) ?

The integral of (9x^3-3x^2+20x+20)/(x^4-16) is

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integral of (9x^3-3x^2+20x+20)/(x^4-16)

\int \frac{9 x ^{3} - 3 x ^{2} + 20 x + 20}{x ^{4} - 16} d x

\frac{9}{4} ln x^{4} - 16 - 2 arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ - \frac{5}{4} ln \frac{x ^{2}}{4} + 1 + \frac{5}{4} ln \frac{x ^{2}}{4} - 1 + C

Solution steps

\int \frac{9 x ^{3} - 3 x ^{2} + 20 x + 20}{x ^{4} - 16} d x

Expand

\frac{9 x ^{3} - 3 x ^{2} + 20 x + 20}{x ^{4} - 16} : \frac{9 x ^{3}}{x ^{4} - 16} - \frac{3 x ^{2}}{x ^{4} - 16} + \frac{20 x}{x ^{4} - 16} + \frac{20}{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{9 x ^{3}}{x ^{4} - 16} d x - \int \frac{3 x ^{2}}{x ^{4} - 16} d x + \int \frac{20 x}{x ^{4} - 16} d x + \int \frac{20}{x ^{4} - 16} d x

\int \frac{9 x ^{3}}{x ^{4} - 16} d x = \frac{9}{4} ln x^{4} - 16

\int \frac{3 x ^{2}}{x ^{4} - 16} d x = 3 (\frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣)

\int \frac{20 x}{x ^{4} - 16} d x = - \frac{5}{4} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1)

\int \frac{20}{x ^{4} - 16} d x = 20 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣)

= \frac{9}{4} ln x^{4} - 16 - 3 (\frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣) - \frac{5}{4} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) + 20 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣)

Simplify

\frac{9}{4} ln x^{4} - 16 - 3 (\frac{1}{4} arctan (\frac{x}{2}) - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣) - \frac{5}{4} (ln \frac{x ^{2}}{4} + 1 - ln \frac{x ^{2}}{4} - 1) + 20 (- \frac{1}{16} arctan (\frac{x}{2}) - \frac{1}{32} ln ∣ x + 2 ∣ + \frac{1}{32} ln ∣ x - 2 ∣) : \frac{9}{4} ln x^{4} - 16 - 2 arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ - \frac{5}{4} ln \frac{x ^{2}}{4} + 1 + \frac{5}{4} ln \frac{x ^{2}}{4} - 1

= \frac{9}{4} ln x^{4} - 16 - 2 arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ - \frac{5}{4} ln \frac{x ^{2}}{4} + 1 + \frac{5}{4} ln \frac{x ^{2}}{4} - 1

Add a constant to the solution

= \frac{9}{4} ln x^{4} - 16 - 2 arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣ - \frac{5}{4} ln \frac{x ^{2}}{4} + 1 + \frac{5}{4} ln \frac{x ^{2}}{4} - 1 + C

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What is the integral of (9x^3-3x^2+20x+20)/(x^4-16) ?
The integral of (9x^3-3x^2+20x+20)/(x^4-16) is