Integration by partial Fractions question

  • Thread starter bns1201
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  • #1
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Homework Statement


The problem asks to evaluate the integral using partial fractions, but I just cannot find out which trick to get this one to work. the equation is
[tex]\int[/tex][tex]\frac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}[/tex]



The Attempt at a Solution



I have tried setting it up as a partial fraction with each part of the numerator under Ax+B and Cx+D but I have a feeling that is incorrect. Thanks in advance!
 

Answers and Replies

  • #2
Dick
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Don't let a feeling it's incorrect stop you from doing it. If you followed through with your original program, you'd be done now. That's perfectly correct.
 
  • #3
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When I match up the coefficients to the problem i then get

A+C=1

B+D=1

2A+C=2

2B+D=1


This is where I get stuck again because how can B+D and 2B+D both equal 1?
 
  • #4
Dick
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They can both be equal to 1 if B=0 and D=1. Don't overthink it. Just solve the equations. Subtract B+D=1 from 2B+D=1 and you'll reach that conclusion.
 
  • #5
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thanks...i always over think the problems and forget about 0....thanks so much. calc 2 is really kickin my butt
 
  • #6
Dick
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You could kick it's butt instead. Everything you posted was exactly correct. Just push ahead and try it before you start thinking you must be wrong.
 
  • #7
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I'm having problems with another integral of similar properties.

The integral is
[tex]\int[/tex][tex]\frac{1}{x\sqrt{x+1}}[/tex]

I made this into a u sub, where u = [tex]\sqrt{x+1}[/tex] and u^2 = x+1 Therefore x+u^2-1


So the integral is now [tex]\int[/tex][tex]\frac{1}{u^3-u}[/tex] which i made into [tex]\int[/tex][tex]\frac{1}{u(u-1)(u+1)}[/tex]

I then used partial fractions to make it [tex]\frac{A}{u}[/tex] + [tex]\frac{B}{u+1}[/tex]+[tex]\frac{C}{u-1}[/tex]


Getting me A=-1 B=[tex]\frac{1}{2}[/tex] and C=[tex]\frac{1}{2}[/tex]

Which changed to equation into [tex]\int[/tex][tex]\frac{-1}{u}[/tex]+[tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{du}{u+1}[/tex]+[tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{du}{u-1}[/tex]


Which gets me to the answer -ln|[tex]\sqrt{x+1}[/tex]|+ 1/2 ln|[tex]\sqrt{x+1}+1[/tex]|+1/2 ln |[tex]\sqrt{x+1}-1[/tex]|

but the answer is ln [tex]\frac{[tex]\sqrt{x+1}+1}{\sqrt{x+1}-1}[/tex]





Can anyone provide me with help as to where I am messing up? Thanks
 
  • #8
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In your original integral you omitted dx. In your substitution, you neglected to calculate du, which is not equal to dx.
 
  • #9
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Thanks, I'll redo the equation right now
EDIT: ahhh I did it over with dx in there....thanks a bunch
 
  • #10
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Another question please:
I have to do

[tex]\int[/tex][tex]\frac{2x}{(x-3)^2}[/tex] with upper bound 2 and lower bound 0

What I've done so far:

Partial Fractions, making it: [tex]\frac{A}{x-3}[/tex]+[tex]\frac{B}{(x-3)^2}[/tex]

=> A(x-3)+B=2x
=> A=2 B=6
=> [tex]\int[/tex][tex]\frac{2}{x-3}[/tex] + [tex]\frac{6}{((x-3)^2)}[/tex]
=> 2 ln|x-3| - 6 ln|t-3| again, evaluated from 0 to 2

I end up with -2 ln|3| + 6 ln|3|

The answer is 4-ln|9| Can anyone tell me where I'm going wrong? Thanks
 
  • #11
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Your partial fraction decomposition was fine, but your 2nd integral wasn't.
[tex]\int \frac{6}{(x - 3)^2}dx \neq ln|(x - 3)^2| + C[/tex]

Also, you could have done the problem more simply by using an ordinary substitution: u = x - 3, du = dx. You would have gotten u + some stuff in the numerator and u^2 in the denominator, which can be integrated easily.
 
  • #12
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Thanks with the hint on u substitution, that was a lot easier


Another question if possible.

How would you start the integral

[tex]\int[/tex][tex]\frac{earctany)}{1+y^2}[/tex]

with boundries of -1 to 1

I can't seem to find the trick...
 
Last edited:
  • #13
35,291
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u = arctan(y).
What is du?

If you omit du, you're hosed.

Hint: pay attention to and don't lose those differentials!
 
  • #14
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ahhh yes i see it now. Thanks a bunch.
 
  • #15
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Can anyone help me as to where to start this one out? Thanks


[tex]\int[/tex](1+[tex]\sqrt{x}[/tex])^8 dx
 
  • #16
Dick
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Start with the obvious and unavoidable u=1+sqrt(x). Where does that take you?
 
Last edited:
  • #17
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thanks again I appreciate it
 

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