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?
 A: You don't really need decomposition into partial fractions. Actually, integrals $\;\displaystyle I_n=\int\frac{\mathrm dt}{(1-t^2)^n} \;$ and $\;\displaystyle J_n=\smash{\int\frac{\mathrm dt}{(1+t^2)^n}} $ are best computed recursively. 
Note that for $n=1$, it is basic that
$$\int\frac{\mathrm dt}{ 1-t^2}=\arg\!\tanh x=\frac12\ln\Bigl(\frac{1+x}{1-x}\Bigr) $$
Now, for the computation of $I_2$, perform integration by parts for $I_1$:
Set $\;u=\dfrac{1}{ 1-t^2},\;\mathrm dv=\mathrm dt$, whence $\;\mathrm du=\dfrac{2t\,\mathrm dt}{(1-t^2)^2},\;v=t$. We obtain
\begin{align}
\int\frac{\mathrm dt}{ 1-t^2}&=\frac{t}{ 1-t^2} -2\int\dfrac{t^2\,\mathrm dt}{(1-t^2)^2}=\frac{t}{ 1-t^2} -2\int\dfrac{(t^2-1+1)\,\mathrm dt}{(1-t^2)^2}\\
&=\frac{t}{ 1-t^2} +2\int\dfrac{\mathrm dt}{1-t^2}-2\int\dfrac{\mathrm dt}{(1-t^2)^2}.
\end{align}
Can you end the computation?
A: Correcting a small error in the problem statement shows that the partial fractions decomposition of the integrand has the form
$$\frac{1}{(t^2 - 1)^2} = \frac{A}{(t - 1)^2} + \frac{B}{t - 1} + \frac{C}{(t + 1)^2} + \frac{D}{t + 1}.$$
Hint Since the left-hand side is even, so is the right-hand side.

 Evenness imposes precisely that $$C = A, D = - B ,$$ leaving a system of two unknowns, $A, B$. There are several options for proceeding. A cheap one is evaluating both sides at $t = 0$, giving $1 = A - B + A - B = 2 (A - B)$.

A: Hint: It must be $$1/4\, \left( t+1 \right) ^{-2}+1/4\, \left( t-1 \right) ^{-2}-1/4\,
 \left( t-1 \right) ^{-1}+1/4\, \left( t+1 \right) ^{-1}
$$
A: 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
$$I_1=\frac{x}{2(x^2-1)}+\operatorname{arctanh}x+C$$
A: $$\frac{1}{t^2-1} = \frac{1}{2}\left(\frac{1}{t-1} - \frac{1}{t+1}\right)\tag{1}$$
Square both sides:
$$\frac{1}{\left(t^2-1\right)^2} = \frac{1}{4}\left(\frac{1}{(t-1)^2} + \frac{1}{(t+1)^2}\right) - \frac{1}{2} \frac{1}{t^2-1}$$
The last term's partial fraction expansion is given by (1), therefore we have:
$$\frac{1}{\left(t^2-1\right)^2} = \frac{1}{4}\left(\frac{1}{(t-1)^2} - \frac{1}{t-1} +\frac{1}{(t+1)^2}+\frac{1}{t+1}\right)$$
A: Integrate directly by parts
$$\int \frac{dt}{(t^2-1)^2}=\int \frac1{2t}d\left( \frac{-t^2}{t^2-1}\right)= -\frac12 \frac{t}{t^2-1}-\frac12\int \frac{dt}{t^2-1}
$$
