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Partial Fractions

  • Thread starter LadiesMan
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[SOLVED] Partial Fractions

1. Evaluate:

[tex]\int \frac{dx}{x^{2} -1}[/tex]



Attempt:

[tex]\int \frac{dx}{x^{2} -1}[/tex]

[tex]= \int \frac{dx}{(x+1)(x-1)}[/tex]

[tex]= \frac{A}{(x+1)} + \frac{B}{(x-1)}[/tex]

[tex]= \frac{Ax - A + Bx + B}{(x+1)(x-1)}[/tex]

Where do I got from here? Thanks
 

Answers and Replies

tiny-tim
Science Advisor
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Hi LadiesMan! :smile:

(erm … what happened to the integral sign, and dx … ?)

Easy! You just solve Ax - A + Bx + B = 1.

So A = … , B = … ? :smile:
 
96
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so A = Ax + Bx + B -1
and B = -Ax + A - Bx -1??
 
rock.freak667
Homework Helper
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so A = Ax + Bx + B -1
and B = -Ax + A - Bx -1??
Ax-A+Bx+B=1 for all values of x

meaning for any value of x you put in there, it will always be equal to 1
Choose some suitable values for x to get A and B
 
458
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You could do A(x-1) + B(x+1) = 1

Then to get rid of the A, substitute x = 1 and find B.

To get rid of the B, substitute x = -1 and find A.
 
96
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That means:

(1/2)ln(x-1) - (1/2)ln(x+1) + C

Therefore Since: (1/2)ln(x-1) = ln(x-1)^1/2 and (1/2)ln(x +1)= (1/2)ln(x +1)^1/2

ln(x-1)^1/2 - (1/2)ln(x +1)^1/2 + C

Natural log rule ln a - ln b = ln (a/b)

Thank you =)
 
tiny-tim
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… never substitute …

Choose some suitable values for x to get A and B
No! :frown:

It's a polynomial, and you just put all the coefficients equal to 0.

It's (A+B)x + (B-A-1) = 0,

so both the brackets must be 0. :smile:

Never substitute - it makes it look as if you don't understand what a polynomial is!

(I agree that, technically, that's the same as substituting x = 0 and ∞; but that requires an extra line, and it doesn't work if you have x^2 or higher.)
 
HallsofIvy
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Tiny Tim, since that equation is true for all x, you certainly could get substitute any two values of x to get 2 equations to solve for a and b. Of course, it is true that if a polynomial is 0 for all x, then its coefficients must all be 0- that's typically simpler than substituting specific values for x.

Ladies Man, a somewhat simpler method for partial fraction is this:
You want to find A, B such that
[tex]\frac{A}{x+1}+ \frac{B}{x-1}= \frac{1}{x^2- 1}[/tex]
Multiply both sides by x2-1 to eliminate the denominators:
[tex]A(x-1)+ B(x+ 1)= 1[/itex]
Now substitute x= 1 and x= -1 to find A and B.
 
96
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Thanks everyone! =P
 

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