Integral $\int \:\frac{dt}{\left(t^2-1\right)^2}$ Compute the following integral:
$$\int \:\frac{dt}{\left(t^2-1\right)^2}$$
So I managed to factor the denominator and put the single fraction into partial fractions:
$$\int \frac{A}{\:t+1}+\frac{B}{\left(t+1\right)^2}+\frac{C}{t-1}+\frac{D}{\left(t-1\right)^2}\,dt$$
But I was a little confused on how to multiply the initial denominator to both sides.
The result is supposed to be:
$1=A\left(t+1\right)\left(t-1\right)^2+B\left(t-1\right)^2+C\left(t-1\right)\left(t+1\right)^2+D\left(t+1\right)^2$
However, I am not exactly sure how the right side of this equation came about.
Any help?
 A: Here is part of the answer:
$$\frac{A}{t+1}\,(t^2-1)^2=\frac{A}{t+1}\,(t+1)^2(t-1)^2=A(t+1)(t-1)^2\ .$$
I'm sure you can do the rest for yourself.
A: We have $$\frac{1}{(t^2-1)^2}=\frac{1}{(t+1)^2(t-1)^2}=\frac{A}{t+1}+\frac{B}{(t+1)^2}+\frac{C}{t-1}+\frac{D}{(t-1)^2}.$$Now just multiply both sides by $(t+1)^2(t-1)^2$ and clear the denominators.
A: You want each denominator of your partial fractions to "match" the initial denominator $(t^2-1)^2=(t+1)(t+1)(t-1)(t-1)$. For the partial fraction with $A$, you already have the factor $t+1$, so you are only missing $(t+1)$, $(t-1)$ and another $(t-1)$, hence you need to "add" $(t+1)(t-1)^2$ to the denominator to make the match. However, if you want to keep the fraction unchanged, you need to do the same to your numerator. This is why you get the term "$A(t+1)(t-1)^2$" in your result.
Repeat the same steps with the other 3 partial fractions and you will get the result you posted (which compares the "initial" numerator, 1, to the one you obtain through partial fractions).
A: Hint
$$I=\int \:\frac{dt}{\left(t^2-1\right)^2}=\frac 12\int \:\frac{(t+1)-(t-1)dt}{\left(t^2-1\right)^2}$$
Simplify both integrals
$$2I=\int \frac {dt} {(t+1)(t-1)^2}-\int \frac {dt} {(t+1)^2(t-1)}$$
Apply the trick again...
