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

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

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

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

\frac{x ^{4}}{4 ( x ^{4} - 5 x ^{2} + 4 )} + \frac{8 x ^{4} ln ∣ x + 2 ∣ + 8 x ^{4} ln ∣ x - 2 ∣ - 2 x ^{4} ln ∣ x + 1 ∣ - 2 x ^{4} ln ∣ x - 1 ∣ + 10 x ^{2} ln ∣ x + 1 ∣ + 10 x ^{2} ln ∣ x - 1 ∣ - 15 x ^{2} - 40 x ^{2} ln ∣ x + 2 ∣ - 40 x ^{2} ln ∣ x - 2 ∣ + 32 ln ∣ x + 2 ∣ + 32 ln ∣ x - 2 ∣ - 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ + 12}{12 ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 )} + 4 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣) + C

Solution steps

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

Expand

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

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

\int \frac{x ^{3}}{x ^{4} - 5 x ^{2} + 4} d x = \frac{x ^{4}}{4 ( x ^{4} - 5 x ^{2} + 4 )} + \frac{8 x ^{4} ln ∣ x + 2 ∣ + 8 x ^{4} ln ∣ x - 2 ∣ - 2 x ^{4} ln ∣ x + 1 ∣ - 2 x ^{4} ln ∣ x - 1 ∣ + 10 x ^{2} ln ∣ x + 1 ∣ + 10 x ^{2} ln ∣ x - 1 ∣ - 15 x ^{2} - 40 x ^{2} ln ∣ x + 2 ∣ - 40 x ^{2} ln ∣ x - 2 ∣ + 32 ln ∣ x + 2 ∣ + 32 ln ∣ x - 2 ∣ - 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ + 12}{12 ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 )}

\int \frac{4}{x ^{4} - 5 x ^{2} + 4} d x = 4 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣)

= \frac{x ^{4}}{4 ( x ^{4} - 5 x ^{2} + 4 )} + \frac{8 x ^{4} ln ∣ x + 2 ∣ + 8 x ^{4} ln ∣ x - 2 ∣ - 2 x ^{4} ln ∣ x + 1 ∣ - 2 x ^{4} ln ∣ x - 1 ∣ + 10 x ^{2} ln ∣ x + 1 ∣ + 10 x ^{2} ln ∣ x - 1 ∣ - 15 x ^{2} - 40 x ^{2} ln ∣ x + 2 ∣ - 40 x ^{2} ln ∣ x - 2 ∣ + 32 ln ∣ x + 2 ∣ + 32 ln ∣ x - 2 ∣ - 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ + 12}{12 ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 )} + 4 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣)

Add a constant to the solution

= \frac{x ^{4}}{4 ( x ^{4} - 5 x ^{2} + 4 )} + \frac{8 x ^{4} ln ∣ x + 2 ∣ + 8 x ^{4} ln ∣ x - 2 ∣ - 2 x ^{4} ln ∣ x + 1 ∣ - 2 x ^{4} ln ∣ x - 1 ∣ + 10 x ^{2} ln ∣ x + 1 ∣ + 10 x ^{2} ln ∣ x - 1 ∣ - 15 x ^{2} - 40 x ^{2} ln ∣ x + 2 ∣ - 40 x ^{2} ln ∣ x - 2 ∣ + 32 ln ∣ x + 2 ∣ + 32 ln ∣ x - 2 ∣ - 8 ln ∣ x + 1 ∣ - 8 ln ∣ x - 1 ∣ + 12}{12 ( x + 1 ) ( x - 1 ) ( x + 2 ) ( x - 2 )} + 4 (\frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} ln ∣ x - 2 ∣) + C

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