<s>[INST] Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$ I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:
$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{2}{(x^2-1)^2}dx+\int\frac{1}{(x^2-1)^3}dx$$
I don't know how to evaluate remaining integrals.
 [/INST] Integrate by parts to reduce the remaining integrals
$$\int \frac{dx}{(x^2-1)^3}=\int \frac1{4x^3}d\left( \frac{-x^4}{(x^2-1)^2}\right)= -\frac14 \frac{x}{(x^2-1)^2}-\frac34 \int \frac{dx}{(x^2-1)^2}
$$
$$\int \frac{dx}{(x^2-1)^2}=\int \frac1{2x}d\left( \frac{-x^2}{x^2-1}\right)= -\frac12 \frac{x}{x^2-1}-\frac12\int \frac{dx}{x^2-1}
$$
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