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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?

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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.

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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.

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  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Clayton Feb 18 '18 at 18:52
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    $\begingroup$ It wasn't my downvote but I would guess that since you have basically just said to multiply the initial denominator to both sides, and the OP asked how to do this, someone thought it didn't seem like much of an answer to the question. $\endgroup$ – David Feb 18 '18 at 23:39
  • $\begingroup$ @David: Perhaps you are correct; I understand it (even now) as the OP asking why multiplying by the initial denominator gives the expansion he is seeking. Also, +$1$ for your answer. It helps balance the inexplicable downvote, as well as answering the question clearly. $\endgroup$ – Clayton Feb 19 '18 at 0:34
  • $\begingroup$ It's pretty usual for downvotes to be unexplained, that's an unfortunate fact of life on this site. I must say I rarely bother asking any more. $\endgroup$ – David Feb 19 '18 at 0:40
  • $\begingroup$ @David it's the way it is here on this site...sometimes they downvotes for arithmetic mistakes...It's really weird...anyway +1 for your answer $\endgroup$ – Isham Feb 19 '18 at 2:36
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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).

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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...

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