= Solution
For a <partial fraction decomposition>, write
$$
\frac4{(x-1)^2(x+1)}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+1}.
$$
Multiplying by $(x-1)^2(x+1)$ gives $4=A(x^2-1)+B(x+1)+C(x-1)^2$. Setting $x=1,-1$ gives $B=2,C=1$, and comparing the coefficient of $x^2$ gives $A=-1$. Thus
$$
f(x)=-\frac1{x-1}+\frac2{(x-1)^2}+\frac1{x+1}.
$$
Integrating each term gives the <antiderivative>
$$
\boxed{\int f(x)\,dx=\log\left|\frac{x+1}{x-1}\right|-\frac2{x-1}+C.}
$$
The <constant of integration> can be chosen independently on each interval separated by $-1$ and $1$.
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