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Thread: 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:

    $\displaystyle \int \frac{x^2-3x+7}{(x^2-4x+6)^2}dx$
    (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:

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

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


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

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

    This is just after i finished making the integral into partial fractions. Also $\displaystyle u=x-2$ 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:

    $\displaystyle \int \frac{x^2-3x+7}{(x^2-4x+6)^2}dx$
    (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:

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

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

    This is just after i finished making the integral into partial fractions. Also $\displaystyle u=x-2$ 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:

    $\displaystyle \int \frac{1}{u^2+2}\,du = \tfrac{1}{2}\int\frac{1}{\left(\tfrac{u}{\sqrt{2}} \right)^2+1}\,du = \ldots$
    As to the second: a simple substitution like $\displaystyle z= u^2+2$ 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 $\displaystyle z=u^2+2$.

    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 $\displaystyle +constant$ 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

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

    so you pretend $\displaystyle 2=a^2$

    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 $\displaystyle 3\int(u^2+2)^{-2}du$and 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 $\displaystyle \int\frac{1}{x}dx$ 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

    $\displaystyle x^2 - 3x + 7 $

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

    we obtain

    $\displaystyle A + C = 1 $ (1)

    $\displaystyle 4A - B = 3 $ (2)

    $\displaystyle 6A - 2B - 6C = 7 $ (3)

    $\displaystyle 2(2) - (3) $ ,

    $\displaystyle 2A + 6C = -1 $ (4)

    $\displaystyle 2(1) - (4) $ ,

    $\displaystyle -4C = 3 $ .....

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



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

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

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

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

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

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


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

    $\displaystyle \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 $\displaystyle = span[ x^2 - 4x + 6 , x-2 , x^2 - 6 ]$
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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 $\displaystyle +constant$ 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

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

    so you pretend $\displaystyle 2=a^2$
    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):

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

    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 $\displaystyle 3\int(u^2+2)^{-2}du$and 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 $\displaystyle \int\frac{1}{x}dx$ 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 $\displaystyle \lambda := \sqrt{2}$, and consider the integral
    $\displaystyle I_n := \int\frac{1}{(u^2+\lambda^2)^n}\,du$
    We have already found that
    $\displaystyle I_1 = \int\frac{1}{(u^2+\lambda^2)^1}\, du =\frac{1}{\lambda}\arctan\frac{u}{\lambda}+C$
    Now, by partial integration, we can derive a recursion formula for the value of $\displaystyle I_n$:
    $\displaystyle I_n = \int \underset{\uparrow}{1}\cdot\underset{\downarrow}{\ frac{1}{(u^2+\lambda^2)^n}}\, du=u\cdot\frac{1}{(u^2+\lambda^2)^n}-\int u\cdot\frac{-n\cdot 2u}{(u^2+\lambda^2)^{n+1}}$
    This last integral seems more complicated than before, but a little trick helps here:
    $\displaystyle =\frac{u}{(u^2+\lambda^2)^n}+2n\int\frac{u^2+\lamb da^2}{(u^2+\lambda^2)^{n+1}}\,du-2n\lambda^2\int \frac{1}{(u^2+\lambda^2)^{n+1}}\, du$
    So, we have
    $\displaystyle I_n=\frac{u}{(u^2+\lambda^2)^n}+2n I_n-2n\lambda^2I_{n+1}$
    Solve for $\displaystyle I_{n+1}$ to get the promised recursion formula and plug in your $\displaystyle \lambda=\sqrt{2}$, and n=2, to get what you need.
    Last edited by Failure; Apr 7th 2010 at 11:08 AM.
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