What is the integral of (x^9)/((x^4-1)^2) ?

The integral of (x^9)/((x^4-1)^2) is

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integral of (x^9)/((x^4-1)^2)

\int \frac{x ^{9}}{( x ^{4} - 1 ) ^{2}} d x

\frac{x ^{2}}{2} - \frac{3}{8} ln x^{2} + 1 - \frac{1}{8 ( x ^{2} + 1 )} + \frac{3}{8} ln ∣ x + 1 ∣ + \frac{1}{16 ( x + 1 )} + \frac{3}{8} ln ∣ x - 1 ∣ - \frac{1}{16 ( x - 1 )} + C

Solution steps

\int \frac{x ^{9}}{( x ^{4} - 1 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{9}}{( x ^{4} - 1 ) ^{2}} : x - \frac{3 x}{4 ( x ^{2} + 1 )} + \frac{x}{4 ( x ^{2} + 1 ) ^{2}} + \frac{3}{8 ( x + 1 )} - \frac{1}{16 ( x + 1 ) ^{2}} + \frac{3}{8 ( x - 1 )} + \frac{1}{16 ( x - 1 ) ^{2}}

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

\int x d x = \frac{x ^{2}}{2}

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

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

\int \frac{3}{8 ( x + 1 )} d x = \frac{3}{8} ln ∣ x + 1 ∣

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

\int \frac{3}{8 ( x - 1 )} d x = \frac{3}{8} ln ∣ x - 1 ∣

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

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

Simplify

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

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{3}{8} ln x^{2} + 1 - \frac{1}{8 ( x ^{2} + 1 )} + \frac{3}{8} ln ∣ x + 1 ∣ + \frac{1}{16 ( x + 1 )} + \frac{3}{8} ln ∣ x - 1 ∣ - \frac{1}{16 ( x - 1 )} + C

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What is the integral of (x^9)/((x^4-1)^2) ?
The integral of (x^9)/((x^4-1)^2) is