<s>[INST] Solving indefinite integral $\int \frac{dx}{(x^4-1)^3}$ I'm trying to solve next integral, but I can't start. WolframAlpha gives me really terrible answer and can't give any step-by-step instructions, so I simply does not know how to start.
$$\int \frac{dx}{(x^4-1)^3}$$
Please give any hint or start point of solving this integral? Maybe there any way to simplify it?
 [/INST] One step that can simplify the partial fraction decomposition is to first perform an integration by parts with the choice $$u = -(2x)^{-3}, \quad du = \frac{3}{8}x^{-4} \, dx, \quad dv = -\frac{(2x)^3}{(x^4-1)^{3}} \, dx, \quad v = (x^4-1)^{-2},$$ giving $$\int \frac{dx}{(x^4-1)^3} = -\frac{1}{8x^3(x^4-1)^2} - \frac{3}{8} \int \frac{dx}{x^4 (x^4-1)^2}.$$  Partial fraction decomposition on $(x^2 (x^4-1))^{-2}$ results in a simpler integrand with lower-degree polynomial denominator terms:
$$\frac{1}{x^4(x^4-1)^2} = \frac{1}{x^4}+\frac{3}{4 \left(x^2+1\right)}+\frac{1}{4(x^2+1)^2}+\frac{7}{16 (x+1)}+\frac{1}{16 (x+1)^2}-\frac{7}{16(x-1)}+\frac{1}{16 (x-1)^2},$$ all of which are trivially integrable with the exception of the third term, which may be handled with the usual trigonometric substitution $x = \tan \theta$, $dx = \sec^2 \theta \, d\theta$, resulting in the transformed integrand $\frac{1}{4} \cos^2 \theta$, which again is easily handled.
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