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Challenged by my teacher: integration by substitution

  1. Feb 26, 2008 #1
    I just learned how to integrate through substitution and I was challenged by my teacher with an apparently easy problem but I'm really struggling with it.
    He said he will give me an F if I don't solve it for tomorrow, I guess this is what I get by being the one who always understand in class

    Exposed below

    1. The problem statement, all variables and given/known data

    [tex]\int (x^2+1)^2 dx[/tex]

    2. Relevant equations




    3. The attempt at a solution

    [tex]\int u^2(\frac{du}{2x}) = [/tex]

    [tex]\int u^2(\frac{du}{2\sqrt{u-1}}) = [/tex]

    [tex]\frac{1}{2}\int\frac{du}{u^{-2}\sqrt{u-1}} = [/tex]

    Now I finally integrate

    \frac{1}{2} ln(u^{-2}\sqrt{u-1})[/tex]

    [tex]\frac{1}{2} ln(\frac{\sqrt{u-1}}{u^2})[/tex]

    [tex]\frac{1}{2} ln(\frac{x}{(x^2+1)^2})[/tex]

    But if I integrate it solving the binomial first I get a totally different answer and If I derive my result I don't get [tex](x^2+1)^2[/tex]

    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2
    You can check your answer by just foiling out [itex](x^2+1)^2[/itex]. So you really shouldn't be getting natural logs...

    Perhaps he's just trying to get you to use u substitution when you don't have to?
  4. Feb 26, 2008 #3
    Yes, that was my first thought but he said that I must use substitution, and this is the only way I think its possible.

    Thanks for your answer!
  5. Feb 26, 2008 #4


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    But, that is wrong. That is not the value of the integral.
    you get a different answer because you get the right answer, which is different from the wrong answer.
  6. Feb 26, 2008 #5
    I'm just learning this, is there anybody out there willing to show me how to integrate it correctly through substitution?
  7. Feb 26, 2008 #6


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    That's clearly not right.
  8. Feb 26, 2008 #7
    Isn't this one of the formulas?

    [tex]\int \frac{dx}{x} = ln(x)[/tex]

    I'm really trying, and I would appreciate answers with more than one line. I know it's wrong but if somebody would be so kind and try to explain it to me?

    I just learned integration three days ago and I learned substitution this morning
    Last edited: Feb 26, 2008
  9. Feb 26, 2008 #8
    Hi Pablo,

    Sorry if it seems that we're being a bit short with our answers. The homework forum isn't a place where others do your homework for you, just offer advice.

    In any case, You're right that [itex]\int \frac{dx}{x} = \ln(x)[/itex] - but this isn't what you have, is it? Instead, you have [itex]\int \frac{dx}{x^{-2}\sqrt{1-x}}[/itex] - that is, [itex]\int \frac{dx}{f(x)} \ne \ln(f(x))[/tex] in general.

    As for how to integrate that thing, I'm afraid I'm stumped without further substitution. Integrals.com and the back of Stewart's calculus book gives [itex]2/30 \sqrt{u-1}(3u^2+4u+8)[/itex] the answer, which, when substituting back in your original substitution gives the correct answer.

    EDIT: Stewart's Calculus has the following:

    [tex]\int \frac{u^2du}{\sqrt{a+bu}}=\frac{2}{15b^3}(8a^2+3b^2u^2-4abu)\sqrt{a+bu}+C[/tex].

    For your integral, [itex]a=b=1[/itex]
    Last edited: Feb 26, 2008
  10. Feb 26, 2008 #9


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    Why would you think of trying to solve that integral by substitution? Any substitution (at least that I can think of) would just make the integrand more complicated. If I were you, I'd just expand and solve it in two lines (of course you can make the substitution u=x if your teacher demands a substitution ;) ) and move on to trying to solve more complicated problems that will help you hone your techniques for the future.
  11. Feb 26, 2008 #10
    Yes I know you are not doing my homework, and it's not really my homework, it's a challenge put by my teacher just to me. I'm also a permanent user in other forums about technical topics and I always try to be as detailed and kind as possible when I answer questions. Even If they are homework I give hints and tips to the poster.

    Mathematica gave me the same answer, thank you however

    Cristo: Yes, it's a lot easier without substitution and I can solve it that way right now, but I'm asked to use substitution. I think I'll stick to u=x like you said. Thanks
  12. Feb 26, 2008 #11
    Thanks for your edit and formula, I think I will use it although I haven't seen that formula on class.
  13. Feb 26, 2008 #12


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    You can't apply that rule -- [itex]du / (u^{-2}\sqrt{u-1})[/itex] simply doesn't match the pattern [itex]dx / x[/itex].
    Last edited: Feb 26, 2008
  14. Feb 26, 2008 #13
    My teacher seems to be a moron, he used the formula a couple of times like I did. I think this led to my confusion. I believe the correct way is with PingPong's formula
  15. Feb 26, 2008 #14
    Thanks Hurkyl PingPong, your help is really appreciated
  16. Feb 27, 2008 #15


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    But the point of substitutions is to make an integral simpler to compute, not make it harder! If your teacher did intend that to be the solution, then I have no idea what he is thinking-- what a waste of time!
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