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Integration of exp(x^2)*(x^2-1)/x^3

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find the integral of [itex]\frac{exp(x^{2})(x^{2}-1)}{x^{3}}[/itex]

    2. Relevant equations
    Wolfram returns this answer
    [itex]\frac{exp(x^{2})}{2x}[/itex]


    3. The attempt at a solution
    I orginally had the integral in the form of:
    [itex]\frac{exp(x^{2})(x^{2})}{x^{3}}[/itex]-[itex]\frac{exp(x^{2})}{x^{3}}[/itex]

    But I dont know how to integrate either!

    Any help or suggestions will be appreciated!
     
  2. jcsd
  3. Oct 1, 2012 #2
    Well, it's not necessary, but you can let u=x2. That should make it a little easier. You should use the chain, product and quotient rules.
     
  4. Oct 1, 2012 #3

    SammyS

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    [itex]\displaystyle \frac{e^{(x^2)}(x^{2}-1)}{x^3}=\frac{e^{(x^2)}x^2}{x^3}-\frac{e^{(x^2)}}{x^3}[/itex]
    [itex]\displaystyle =
    \frac{e^{(x^2)}}{x}-\frac{e^{(x^2)}}{x^3}
    [/itex]​
    The substitution suggested by hogrampage is the place to start.

    Then it looks like you will need to do integration by parts.

    Added in Edit:

    Well, yes, it can be done by inspection using the quotient rule in 'reverse'. Simply multiply the integrand by x/x.

    [itex]\displaystyle =
    \frac{1}{2}\frac{2xe^{(x^2)}x^{2}-e^{(x^2)}\,2x}{x^4}
    [/itex]

    I presume it can be done similarly using the product rule.
     
    Last edited: Oct 1, 2012
  5. Oct 5, 2012 #4
    Thanks!
    I know the answer is [itex]\frac{exp(x)}{2x}[/itex], becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?
     
  6. Oct 5, 2012 #5

    SammyS

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    Actually, that's not the result.

    [itex]\displaystyle \frac{d}{dx}\left(\frac{\exp(x)}{2x}\right)=\frac{e^x (x-1)}{2 x^2}[/itex]

    Earlier you said the result was [itex]\displaystyle \frac{\exp(x^2)}{2x}\ .[/itex]

    That's not it either.

    [itex]\displaystyle \frac{d}{dx}\left(\frac{\exp(x^2)}{2x}\right)= \frac{\exp(x^2)(2x^2-1)}{2x^2}[/itex]

    Do you know integration by parts?
     
  7. Oct 7, 2012 #6
    [tex]\int \frac{e^{x^2}(x^2-1)}{x^4} xdx[/tex] using the substitution suggested I got, [tex]\int \frac{e^{u}(u-1)}{2u^2} du[/tex]= [tex] \frac{e^{u}-1}{2u} [/tex].
    Substituting back to x, [tex] \frac{e^{x^2}-1}{2x^2} [/tex]. The limits on the integration were from 0 to 2.
    [tex] \frac{e^{4}-1}{8} - 0 = 6.6997... [/tex]
    However, when I do this on wolfram, I get, [tex] \frac{e^{4}-5}{8} = 6.1997... [/tex]
    I can't see what I am doing wrong.
     
  8. Oct 7, 2012 #7

    SammyS

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

    [itex]\displaystyle \frac{e^{u}(u-1)}{u^2}=\frac{u\cdot e^{u}-e^{u}\cdot 1}{u^2}[/itex]

    Quotient rule:

    [itex]\displaystyle \frac{\text{(Low) D(High)} - \text{(High) D(Low)}}{\text{(Low)}\!^2}[/itex]
     
  9. Oct 8, 2012 #8
    Thanks Sammy,
    Now I see that I made a mistake at the very beginning.
    I wanted to integrate [tex]\int \frac{e^{x^2}(x^2-1)+1}{x^4} xdx[/tex] in which case I think the integral is correct, and still unsure where the 5 comes from?
     
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