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Homework Help: Hard integral (for me)

  1. Jan 4, 2006 #1

    could you please help me in evaluating this integral?

    \int 3e^{2x}\sqrt{x+1}\ dx

    I got to this point during solving one ODE and have no idea what to do with this one.

    Thank you.
  2. jcsd
  3. Jan 4, 2006 #2


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    not pretty, Maple says it involves an erfc.
  4. Jan 4, 2006 #3
    Bad, very bad...
  5. Jan 4, 2006 #4
    Hi twoflower

    I solve your integral, and found the following answer



    Erfi[z] gives the imaginary error function erf(iz)/i
  6. Jan 4, 2006 #5
    Thank you, but that's definitely not what I was supposed to get..Kind of ugly for ODE solution, isn't it? :)
  7. Jan 4, 2006 #6
    try [tex]y=\sqrt{x+1}[/tex]
    and then pick [tex]4ye^{2y^2}[/tex] as dv and [tex]y[/tex] for u... (http://mathworld.wolfram.com/IntegrationbyParts.html" [Broken]
    Last edited by a moderator: May 2, 2017
  8. Jan 4, 2006 #7


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    Last edited: Jan 4, 2006
  9. Jan 4, 2006 #8
    Sorry, I'm learning to use this editor, the answer is as follow

  10. Jan 4, 2006 #9
    oh yeah, and i forgot, after the integration in parts i suggested above.. you know the trick for [tex]e^{x^2}[/tex] right?
  11. Jan 4, 2006 #10
    Thank you fargoth, gonna try that.

    The ODE is

    y'' + 2y' + y = 3e^{x}\sqrt{x+1}
  12. Jan 4, 2006 #11


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    is it [itex]e^{2x}\mbox{ or }e^{x}[/itex]?
  13. Jan 4, 2006 #12
    It's [itex]e^{x}[/itex]
  14. Jan 4, 2006 #13


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    Well, if it is either of the above, then your answer is just nasty (erfc, and all that). But if, perhaps, you had, say [tex]y^{\prime\prime}+2y^{\prime} + y= 3e^{-x}\sqrt{x+1} [/tex], then your answer is rather sweet, namely [tex]y(x)=c_{1}e^{-x}+c_{2}xe^{-x}+\frac{4}{5}e^{-x}(x+1)^{\frac{5}{2}}[/tex]. In fact, if the exponent of e in the nonhomogeneous part is anything other than -1, erfc (computer aided educated guess).
    Last edited: Jan 4, 2006
  15. Jan 4, 2006 #14
    You're an oracle, benorin! Professor wrote in wrongly, there has to be [itex]e^{-x}[/itex] in the original ODE.
  16. Jan 4, 2006 #15


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    Why should the mildly messy format of the first solution discourage you? The ODE as first written is perfectly solvable even for the second solution (assuming you're using variation of parameter and B(x) is defined as above) with the second one:

    [tex]A(x)=-3\int xe^{2x}\sqrt{x+1}[/tex]

    That's just me though.:smile:

    Edit: Oh yea, I haven't (yet) checked this.

    Edit2: Oh yea: equal rights for special functions. You know I have to send a dollar to Lurflurf each time I use that saying don't you?
    Last edited: Jan 4, 2006
  17. Jan 4, 2006 #16
    Now I computed it and got another solution..

    And the official one our proffesor has on his webpage is

    y = \frac{3}{2} \log \left(x^2 + \sqrt{x^4 + 1}\right)e^{-x} -2(x+1)^{\frac{3}{2}}xe^{-x} + \alpha e^{-x} + \beta xe^{-x}
  18. Jan 4, 2006 #17
    What?? :biggrin:
  19. Jan 4, 2006 #18


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    Hey Twoflower, what's going on with this? Your initial ODE was:


    Using variation of parameter, I get:





    Back-substituting y(x) (via Mathematica) into the LHS of the ODE, I get the RHS.

    When I back-substitute the "official solution" you posted above into the LHS of the ODE, I get an expression not even close to the RHS, minus x or not.
    Last edited: Jan 4, 2006
  20. Jan 5, 2006 #19


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    Really, just what is anyway:




    And also, I think we should ask Twoflower to solve this for us:

    [tex]y^{''}+2y^{'}+y=3e^x\sqrt{1+x},\quad y(0)=0,\;y'(0)=1[/tex]

    not numerically neither: figure out what c1 and c2 are.:smile:
    Last edited: Jan 5, 2006
  21. Jan 7, 2006 #20
    Well, this will definitely not appear in the test and I would probably not manage it.
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