İntegral of Rational Function

[Arman777]

Homework Statement

$∫\frac {dx}{(x^2-1)^2}$

Homework Equations

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 don't know how to do ? Maybe I can just left at simpler terms.But how can I decide it ?

 

[PeroK]

Homework Statement

$∫\frac {dx}{(x^2-1)^2}$

Homework Equations

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 don't know how to do ? Maybe I can just left at simpler terms.But how can I decide it ?

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.

 

[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.

 

[Ray Vickson]

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.

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.

 

[Mark44]

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

 

[PeroK]

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

There is another solution!

 

[Arman777]

There is another solution!

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

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

Is it make things easier ? (I am not sure..)

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.

can lead us yes but too long...

 

[PeroK]

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

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.

 

[Arman777]

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.

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

 

[Ray Vickson]

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...

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.

 

[Arman777]

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

I ll find soon

 

[Arman777]

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.

yeah I know..But they shouldn't made such long questions

 

[Mark44]

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.

yeah I know..But they shouldn't made such long questions

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.

 

[Arman777]

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.

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

 

[PeroK]

yeah I know..But they shouldn't made such long questions

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

 

[Arman777]

I tried every method that we thought but nothing came up

 

[Ray Vickson]

I tried every method that we thought but nothing came up

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

 

[Arman777]

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

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