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Math Help - Crazy Integral Question. Please Help!!!

  1. #1
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    Exclamation Crazy Integral Question. Please Help!!!

    The initial question I have is to evaluate the following indefinite integral:

    \int \frac{x^2-3x+7}{(x^2-4x+6)^2}dx
    (from: calculus, early transcendentals textbook by James Stewart 6th custom edition for MA110. Question from: sec. 7.4 question #37)

    The question seems to require some completing the square of the denominator, substituition, and partial fraction integration. I thought I was on the right track although I have been trying for about 5 hours now and I can't get the same answer as the book which is really frustrating.

    The answer according to my book is:

    \frac{7}{8}\sqrt{2}\arctan(\frac{x-2}{\sqrt{2}})+\frac{3x-8}{4(x^2-4x+6)}+ constant

    If anyone has some time then I would really appreciate the help.


    NOTE: The step I got to and think I am right is:

    Integral = \int\frac{1}{u^2+2}du+\int\frac{u}{(u^2+2)^2}du+\i  nt\frac{3}{(u^2+2)^2}du

    This is just after i finished making the integral into partial fractions. Also u=x-2 if you were wondering what I used as my substitution.

    So ya I got a little stuck here. I have tried finding the integrals seperately although I can't seem to get it to work and it is hard to see where my errors are because the integrals are all seperate and I can't know if even 1 is right. My work so far may also be wrong too though

    Thanks in advance for your help,
    Luke
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Landyach View Post
    The initial question I have is to evaluate the following indefinite integral:

    \int \frac{x^2-3x+7}{(x^2-4x+6)^2}dx
    (from: calculus, early transcendentals textbook by James Stewart 6th custom edition for MA110. Question from: sec. 7.4 question #37)

    The question seems to require some completing the square of the denominator, substituition, and partial fraction integration. I thought I was on the right track although I have been trying for about 5 hours now and I can't get the same answer as the book which is really frustrating.

    The answer according to my book is:

    \frac{7}{8}\sqrt{2}\arctan(\frac{x-2}{\sqrt{2}})+\frac{3x-8}{4(x^2-4x+6)}+ constant
    But beware: "The Answer" is only defined up to a constant, as wou wrote, and equivalence transformations of the entire term. It is sometimes not at all obvious at first glance that two such terms are, indeed, equivalent up to a constant...

    If anyone has some time then I would really appreciate the help.


    NOTE: The step I got to and think I am right is:

    Integral = \int\frac{1}{u^2+2}du+\int\frac{u}{(u^2+2)^2}du+\i  nt\frac{3}{(u^2+2)^2}du

    This is just after i finished making the integral into partial fractions. Also u=x-2 if you were wondering what I used as my substitution.
    This is ok.

    So ya I got a little stuck here. I have tried finding the integrals seperately although I can't seem to get it to work and it is hard to see where my errors are because the integrals are all seperate and I can't know if even 1 is right. My work so far may also be wrong too though

    Thanks in advance for your help,
    Luke
    So where, exactly, is your problem? As to the first integral:

    \int \frac{1}{u^2+2}\,du = \tfrac{1}{2}\int\frac{1}{\left(\tfrac{u}{\sqrt{2}}  \right)^2+1}\,du = \ldots
    As to the second: a simple substitution like z= u^2+2 will do. And as to the third, I would apply a single step of partial integration to get an u into the numerator. Once you have that you can, again, apply the substitution z=u^2+2.

    But please: Consider first the possibility that your solution is really, if looked at closely enough, equivalent to "The Answer" that your book gives - up to a constant.
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  3. #3
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    Thanks for your help. Also I think you misunderstood my +constant thing. That is simply added because when you find the antiderivative of something it could be whatever integral/antiderivative you found plus any constant C because when you take the derivative the derivative of any constant = 0, therefore you still get the integral you started with. In other words: just ignore the +C.

    The second integral of my 3 I see that I did correctly because I did what you described. However for the first one I'm not sure what you were getting at.

    What I was trying with it was

    \int\frac{1}{u^2+2}du=\frac{1}{\sqrt{2}}\arctan(\f  rac{x-2}{\sqrt{2}})
    since
    \int\frac{dx}{x^2+a^2}=\frac{1}{a}\arctan(\frac{x}  {a})

    so you pretend 2=a^2

    Of course I'm not sure if this worked properly for me but I don't see a reason why it would be wrong. I figure it is probably on the right track since there is an arctan function in the answer.


    As for the last integral you lost me. were you saying something like 3\int(u^2+2)^{-2}duand work from there??? I've tried this but I'm not sure whether or not to use a ln function... because it is not exactly a \int\frac{1}{x}dx integral...(i dont think) but each time I go for an answer by simply working backwards through product rule I can't differentiate it back to its original form (I know this part is probably simple but I completely forget how to do it).
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  4. #4
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    Have you learnt vector space ? But even you not , i think you will understand what i am going to say .

    Write

      x^2 - 3x + 7

     = A(x^2- 4x + 6) + B(x-2) + C(x^2 - 6) where  A,B,C are undetermined coefficients. Later , you will know why i do it in this way .

    we obtain

     A + C = 1 (1)

     4A - B = 3 (2)

     6A - 2B - 6C = 7 (3)

     2(2) - (3) ,

      2A + 6C = -1 (4)

     2(1) - (4)  ,

     -4C = 3 .....

     (A,B,C) = ( \frac{7}{4} , 4 , -\frac{3}{4} )



    For the integral  \int \frac{x^2- 4x + 6}{ (x^2- 4x + 6)^2} ~dx = \int \frac{dx}{ x^2- 4x + 6 } , you know , by substitution .

    For the integral  \int \frac{x-2}{ (x^2- 4x + 6)^2} ~dx you also know , by substituting  x^2 - 4x + 6 = u

    and the last integral  \int \frac{ x^2 - 6}{(x^2 - 4x + 6)^2}~dx , trick is here . it is equal to :

     \int \frac{ 1 - 6/x^2 }{ (x - 4 + 6/x)^2}~dx

    then sub.  x + 6/x = u  ~\implies  (1 - 6/x^2 )dx=du

    the integral is reduced to  \int \frac{du}{(u-4)^2}


    so the first step to this problem is : rewrite the numerator as

      \frac{7}{4}(x^2- 4x + 6) + 4(x-2) - \frac{3}{4}(x^2 - 6)


    By using the words in vector space , i say every quadratic polynomial  = span[ x^2 - 4x + 6 , x-2 , x^2 - 6 ]
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Landyach View Post
    Thanks for your help. Also I think you misunderstood my +constant thing. That is simply added because when you find the antiderivative of something it could be whatever integral/antiderivative you found plus any constant C because when you take the derivative the derivative of any constant = 0, therefore you still get the integral you started with. In other words: just ignore the +C.

    The second integral of my 3 I see that I did correctly because I did what you described. However for the first one I'm not sure what you were getting at.

    What I was trying with it was

    \int\frac{1}{u^2+2}du=\frac{1}{\sqrt{2}}\arctan(\f  rac{{\color{green}x}{\color{red}-2}}{\sqrt{2}})
    since
    \int\frac{dx}{x^2+a^2}=\frac{1}{a}\arctan(\frac{x}  {a})

    so you pretend 2=a^2
    We seem to almost agree on this one. Almost, but not quite. What I suggest is the following (to be more precise than the first time around):

    \int\frac{1}{u^2+2}du <br />
= \tfrac{1}{\sqrt{2}}\cdot \int \frac{1}{\left(\frac{u}{\sqrt{2}}\right)^2+1}\, \frac{du}{\sqrt{2}}=\tfrac{1}{\sqrt{2}}\int \frac{1}{z^2+1}\, dz
    where z := \frac{u}{\sqrt{2}}. The last integral is (known, I assume) to be
    =\tfrac{1}{\sqrt{2}}\arctan z+C=\tfrac{1}{\sqrt{2}}\arctan\frac{u}{\sqrt{2}}+C

    Of course I'm not sure if this worked properly for me but I don't see a reason why it would be wrong. I figure it is probably on the right track since there is an arctan function in the answer.


    As for the last integral you lost me. were you saying something like 3\int(u^2+2)^{-2}duand work from there??? I've tried this but I'm not sure whether or not to use a ln function... because it is not exactly a \int\frac{1}{x}dx integral...(i dont think) but each time I go for an answer by simply working backwards through product rule I can't differentiate it back to its original form (I know this part is probably simple but I completely forget how to do it).
    My idea for the third integral was only almost ok, but not quite. To make things a little more general, let \lambda := \sqrt{2}, and consider the integral
    I_n := \int\frac{1}{(u^2+\lambda^2)^n}\,du
    We have already found that
    I_1 = \int\frac{1}{(u^2+\lambda^2)^1}\, du =\frac{1}{\lambda}\arctan\frac{u}{\lambda}+C
    Now, by partial integration, we can derive a recursion formula for the value of I_n:
    I_n = \int \underset{\uparrow}{1}\cdot\underset{\downarrow}{\  frac{1}{(u^2+\lambda^2)^n}}\, du=u\cdot\frac{1}{(u^2+\lambda^2)^n}-\int u\cdot\frac{-n\cdot 2u}{(u^2+\lambda^2)^{n+1}}
    This last integral seems more complicated than before, but a little trick helps here:
    =\frac{u}{(u^2+\lambda^2)^n}+2n\int\frac{u^2+\lamb  da^2}{(u^2+\lambda^2)^{n+1}}\,du-2n\lambda^2\int \frac{1}{(u^2+\lambda^2)^{n+1}}\, du
    So, we have
    I_n=\frac{u}{(u^2+\lambda^2)^n}+2n I_n-2n\lambda^2I_{n+1}
    Solve for I_{n+1} to get the promised recursion formula and plug in your \lambda=\sqrt{2}, and n=2, to get what you need.
    Last edited by Failure; April 7th 2010 at 12:08 PM.
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