Evaluate $\int \frac{dt}{(t^2-1)^2}$ I have this problem of evaluating the indefinite integral of a rational function
$$\int \frac{dt}{(t^2-1)^2}$$
and I'm a bit unsure about how to proceed. I could use partial fractions:
$$\frac{1}{(t+1)(t-1)(t+1)(t-1)}$$
$$\frac{1}{(t+1)^2(t-1)^2}$$
$$\frac{A}{t+1} + \frac{B}{t-1} + \frac{C}{t+1} + \frac{D}{(t+1)^2}$$
and go from there... but that seems long and complicated. Is there a trick I'm missing?

You may find this helpful. Define
$$I_n=\int\frac{dx}{(ax^2+b)^{n+1}}$$
then integrate by parts with $dv=dx$:
$$I_n=\frac{x}{(ax^2+b)^{n+1}}+2(n+1)\int\frac{ax^2}{(ax^2+b)^{n+2}}dx$$
$$I_n=\frac{x}{(ax^2+b)^{n+1}}+2(n+1)\int\frac{ax^2+b}{(ax^2+b)^{n+2}}dx-2b(n+1)\int\frac{dx}{(ax^2+b)^{n+2}}$$
$$I_n=\frac{x}{(ax^2+b)^{n+1}}+2(n+1)I_{n}-2b(n+1)I_{n+1}$$
We then solve for $I_{n+1}$:
$$I_{n+1}=\frac{x}{2b(n+1)(ax^2+b)^{n+1}}+\frac{2n+1}{2b(n+1)}I_n$$
Then replace $n+1$ with $n$ to get
$$I_n=\frac{x}{2bn(ax^2+b)^n}+\frac{2n-1}{2bn}I_{n-1}$$
Which is a recurrence with the base case $$I_0=\frac1{\sqrt{ab}}\arctan\left[x\sqrt{\frac{a}b}\right]$$
Your integral is given by $n=1,\, a=1,\, b=-1$. For the base case, we have
$$I_0=\frac1{\sqrt{-1}}\arctan x\sqrt{-1}=-i\arctan ix=\operatorname{arctanh}x$$
Thus
$$I_1=\frac{x}{2(x^2-1)}+\operatorname{arctanh}x+C$$
