How to integrate $\int\frac{1}{t^4(1-t^2)}dt$ The problem is:
$$\int\frac{1}{t^4(1-t^2)}dt$$
It is obvious that one needs to use partial decomposition here. However, the problem is the fourth power of t in this equation. I've tried with:
$$\frac{A}{t^4} + \frac{Bt+C}{1-t^2}$$
That does not give me the right answer.
Can someone tell me where my mistake is?
Every help is appreciated.
 A: Method used in other answers are for problems which can't be easily broken into partial fraction.   
$\cfrac{1-t^2 + t^2}{t^4(1-t^2)} = \cfrac{1}{t^4} + \cfrac{1}{t^2(1-t^2)} = \cfrac{1}{t^4} + \cfrac{1 - t^2 + t^2}{t^2(1-t^2)} = \cfrac{1}{t^4} + \cfrac{1}{t^2} + \cfrac{1}{1-t^2}$
A: use that $$\frac{1}{t^4(1-t^2)}=1/2\, \left( t+1 \right) ^{-1}+{t}^{-4}-1/2\, \left( t-1 \right) ^{-1}
+{t}^{-2}
$$ and your integral is given by $$-\frac{1}{3 t^3}-\frac{1}{t}-\frac{1}{2} \log
   (1-t)+\frac{1}{2} \log (t+1)+C$$ make the ansatz $$\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t^3}+\frac{D}{t^4}+\frac{E}{1-t}+\frac{F}{1+t}$$
A: What you should have tried was$$\frac A{t^4}+\frac B{t^3}+\frac C{t^2}+\frac Dt+\frac{Et+F}{1-t^2}.$$
A: Since the fraction is even try with
$$\frac{A}{t^2}+\frac{B}{t^4}+\frac{C}{1-t^2}$$
A: We have $$\frac1{t^4(1-t^2)}=\frac A{t^4}+\frac B{t^3}+\frac C{t^2}+\frac Dt+\frac E{1-t}+\frac F{1+t}$$ so $$1=A(1-t^2)+Bt(1-t^2)+Ct^2(1-t^2)+Dt^3(1-t^2)+Et^4(1+t)+Ft^4(1-t)$$ When $t=0$, we have $1=A$.
When $t=-1$, we have $1=2F\implies F=\dfrac12$.
When $t=1$, we have $1=2E\implies E=\dfrac12$.
Since the function is even, $B=D=0$
Using all of this information, trying with $t=\dfrac12$ gives $C=1$.
Now you can easily integrate.
A: When attempting partial fractions, you should start with the most general case. If the denominator has order 4, the highest order the numerator can have is 3 (for a reduced fraction).
From that intuition, your guess should look like this
$$ \frac{At^3 + Bt^2 + Ct + D}{t^4} + \frac{E}{t-1} + \frac{F}{t+1} $$
which reduces to
$$ \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t^3} + \frac{D}{t^4} + \frac{E}{t-1} + \frac{F}{t+1} $$
Also note that $t^2-1=(t-1)(t+1)$
A: $$\int\frac{1}{t^4(1-t^2)}dt$$
Due to the repeated factors....
$$I=\int\frac{1}{t^4(1-t^2)}dt=\int\frac{t^2+1-t^2}{t^4(1-t^2)}dt$$
$$I=\int \frac {dt}{t^4} +\int\frac{1-t^2+t^2}{t^2(1-t^2)}dt$$
$$I=\int \frac {dt}{t^4} +\int \frac {dt}{t^2}+\int\frac{1}{(1-t^2)}dt$$
A: If we put $t^2=x $, then
$$\frac {1}{x^2 (1-x)}=\frac {A}{x}+\frac {B}{x^2}+\frac{C}{1-x} $$
It becomes easy to find
$$C=1$$
$$B=1$$
and
multiplying by $x $ and $x\to \infty $ gives
$$0=A-C $$
From here, the primitive is for you.
