# İntegral of Rational Function

1. Feb 25, 2017

### Arman777

1. The problem statement, all variables and given/known data
$∫\frac {dx}{(x^2-1)^2}$

2. Relevant equations

3. The attempt at a solution
I tried to divide $\frac {1} {(x^2-1)^2}$ as $\frac {Ax+B} {(x^2-1)} +\frac {Cx^3+Dx^2+Ex+F} {(x^2-1)^2}$

but this looks so complex..I dont know how to do ? Maybe I can just left at simpler terms.But how can I decide it ?

2. Feb 25, 2017

### PeroK

One trick is to just use a single label for a polynomial:

$\frac {1} {(x^2-1)^2} = \frac {A(x)} {(x^2-1)} +\frac {B(x)} {(x^2-1)^2}$

And see what that leads to.

3. Feb 25, 2017

### alan2

Is this specifically meant to be done by partial fractions? I would try a trig substitution first. Didn't do it so I don't know what happens.

4. Feb 25, 2017

### Ray Vickson

Doesn't it lead to $A(x) =0$ and $B(x) =1?$

However, the OP could certainly get
$$\frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1},$$
and thus get
$$\frac{1}{(x^2-1)^2} = \frac{A^2}{(x-1)^2}+ \frac{2 A B}{(x-1)(x+1)} + \frac{B^2}{(x+1)^2},$$
then do something with that.

5. Feb 25, 2017

### Staff: Mentor

If the integral is meant to be evaluated using partial fractions, I would decompose the integrand like this:
$\frac 1 {(x^2-1)^2} = \frac 1 {(x - 1)^2(x + 1)^2} = \frac A {x - 1} +\frac B {(x - 1)^2}+ \frac C {x + 1} + \frac D {(x + 1)^2}$

6. Feb 25, 2017

### PeroK

There is another solution!

7. Feb 25, 2017

### Arman777

I get $A(x)=0$
$B(x)-2A(x)=0$
$A(x)-B(x)=1$
whats the other solution ?

Is it make things easier ? (I am not sure..)
can lead us yes but too long...

8. Feb 25, 2017

### PeroK

The other solution is fairly obvious, but it doesn't lead to a great improvement. You shouldn't have been so hasty to dismiss @Mark44's suggestion. I think your definition of what is "too complicated" is no longer appropriate for the level of mathematics you are dealing with. You shouldn't be baulking at only four coefficients.

9. Feb 25, 2017

### Arman777

Its not too hard to find a b c d but after that.We will left another $(x-1)^2$ and $(x+1)^2$ so I have to do all stuff again..Fine I ll try

10. Feb 25, 2017

### Ray Vickson

Not too long at all. Sometimes things need considerable work, and there is no way to avoid doing it. Sometimes there are just no shortcuts.

11. Feb 25, 2017

### Arman777

The answer is $\frac {1} {4} ln (\frac {x+1} {x-1}) - \frac{x} {2(x^2-1)}+C$

I ll find soon

12. Feb 25, 2017

### Arman777

yeah I know..But they shouldnt made such long questions

13. Feb 25, 2017

### Staff: Mentor

Real life isn't always about short questions.

You never told us whether you had to use a specific technique (such as by using partial fraction decomposition). Another approach that was already mentioned was trig substitution.

14. Feb 25, 2017

### Arman777

You are right.The question is in the partial fraction section

15. Feb 25, 2017

### PeroK

"Nature laughs at the difficulties of integration." - Laplace

16. Feb 25, 2017

### Arman777

I tried every method that we thought but nothing came up

17. Feb 25, 2017

### Ray Vickson

What partial fraction representation are you trying to use? Which parts yield "nothing"?

18. Feb 26, 2017

### Arman777

Its too long to write down here :/.I can take photo maybe..Or I ll ask just my prof more simple