How to integrate $\int\frac{1}{t^4(1-t^2)}dt$

Question

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.

Answer 1

Method used in other answers are for problems which can't be easily broken into partial fraction.

$\cfrac{1-t^2 + t^2}{t^4(1-t^2)} = \cfrac{1}{t^4} + \cfrac{1}{t^2(1-t^2)} = \cfrac{1}{t^4} + \cfrac{1 - t^2 + t^2}{t^2(1-t^2)} = \cfrac{1}{t^4} + \cfrac{1}{t^2} + \cfrac{1}{1-t^2}$

Answer 2

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 $$\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t^3}+\frac{D}{t^4}+\frac{E}{1-t}+\frac{F}{1+t}$$

Answer 3

What you should have tried was$$\frac A{t^4}+\frac B{t^3}+\frac C{t^2}+\frac Dt+\frac{Et+F}{1-t^2}.$$

Answer 4

Since the fraction is even try with

$$\frac{A}{t^2}+\frac{B}{t^4}+\frac{C}{1-t^2}$$

Answer 5

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.

Answer 6

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 $t^2-1=(t-1)(t+1)$

Answer 7

$$\int\frac{1}{t^4(1-t^2)}dt$$ Due to the repeated factors.... $$I=\int\frac{1}{t^4(1-t^2)}dt=\int\frac{t^2+1-t^2}{t^4(1-t^2)}dt$$ $$I=\int \frac {dt}{t^4} +\int\frac{1-t^2+t^2}{t^2(1-t^2)}dt$$ $$I=\int \frac {dt}{t^4} +\int \frac {dt}{t^2}+\int\frac{1}{(1-t^2)}dt$$

Answer 8

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.