# İntegral of Rational Function

Gold Member

## Homework Statement

##∫\frac {dx}{(x^2-1)^2}##

## 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 ?

PeroK
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2020 Award

## Homework Statement

##∫\frac {dx}{(x^2-1)^2}##

## 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 ?

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.

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
Homework Helper
Dearly Missed
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
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} ##

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

There is another solution!

Gold Member
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
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2020 Award
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.

Gold Member
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
Homework Helper
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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.

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

I ll find soon

Gold Member
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 shouldnt made such long questions

Mark44
Mentor
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 shouldnt 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.

Gold Member
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
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2020 Award
yeah I know..But they shouldnt made such long questions

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

• Arman777
Gold Member
I tried every method that we thought but nothing came up

Ray Vickson