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.

If the numerator is $x^{2r},$
$$x^{2r}=(1+x^2-1)^r=?$$
Had the numerator been $x^{2r+1},$
$x^{2r+1}=x(1+x^2-1)^r$ first substitute $x^2-1=y$
Now for $S=\dfrac{d(x/(x^2-1)^n)}{dx},$
$S=1/(x^2-1)^n-2(n+1)x^2(x^2-1)^{n+1}$
Write $x^2=x^2-1+1$ and integrate both sides to find
$$\dfrac x{(x^2-1)^n}=I(n)-2(n+1)(I (n) - I(n+1))$$ where $I(m)=\displaystyle\int\dfrac{dx}{(x^2-1)^m}$
Can you identify the values of  $n$ here?
