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Integration problem

  1. Jul 22, 2013 #1
    1. The problem statement, all variables and given/known data
    find the definite integral [itex]\int\frac{x^3}{\sqrt{x^2 + 1}}[/itex] dx from 0 to 1


    2. Relevant equations



    3. The attempt at a solution
    This problem is in the integration by parts section .. I chose u = x^3 , and dv=[itex]\frac{1}{\sqrt{x^2 + 1}}[/itex] so v = [itex]\frac{x^4}{4}[/itex] and du = -(x^2 + 1)^([itex]\frac{-3}{2}[/itex]) , so the integral is equal to [itex]\frac{x^4}{4}[/itex] . [itex]\frac{1}{√x^2 + 1}[/itex] - [itex]\int\frac{x^4}{4}[/itex] . -(x^2 + 1)^([itex]\frac{-3}{2}[/itex]) .. and my problem is the integral on the right hand side of the equation; I don't know how to integrate it and I don't know whether if I've chosen the parts correctly or there is a better way of choosing the parts
     
    Last edited by a moderator: Jul 22, 2013
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  3. Jul 22, 2013 #2

    hilbert2

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    How about first doing a change of variables s=x2. Now ds=2xdx and the integrand simplifies considerably.
     
  4. Jul 22, 2013 #3

    Zondrina

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    If the change of variable doesn't float your boat, you could use a trig substitution of ##x = atanθ##.
     
  5. Jul 22, 2013 #4

    LCKurtz

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    It isn't the easiest way to work the problem, but if you really want to use integration by parts, try$$
    u = x^2,\,dv=\frac x {\sqrt{x^2+1}}$$
     
  6. Jul 22, 2013 #5

    SteamKing

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    In the OP, if u = x^3, then du = 3x^2 dx, not that weird expression. This is basic differentiation.

    And if dv = dx/SQRT(x^2+1), then v certainly is not equal to (1/4)*x^4.

    You should review the formulas for derivatives and simple integrals.
     
  7. Jul 22, 2013 #6

    HallsofIvy

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    You have "u" and "v" confused. With u= x^3, du= 3x^2dx- you differentiated v. With [itex]dv= \frac{1}{\sqrt{x^2+ 1}}[/itex], [itex]v= arctan(x)[/itex]- you integrated u.

     
  8. Jul 24, 2013 #7
    A one solve the definite integral

    ∫x2/√x2+1 dx={z=x2+1,dz=2*x*dx}=
    ∫(z-1)/(2√z) dz=1/3*z3/2-√z={x=0⇔z=1,x=1⇔z=2}=(2-√2)/3
     
  9. Jul 24, 2013 #8

    LCKurtz

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    All these responses and suggestions, and Zerkor never returned to his thread. :frown:
     
  10. Jul 24, 2013 #9
    I tried this one and it didn't work .. In the sections I've studied in the previous two days I learned trigonometric substitution and miscellaneous substitution; I think I have to try working this problem with those techniques
     
  11. Jul 24, 2013 #10
    Thanks a lot guys for your help and I'm sorry for replying after a long period but I didn't access the forum through those two days :)
     
  12. Jul 24, 2013 #11

    SteamKing

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    Without showing your work, we can't say why integration by parts didn't work for you, except you must have made a mistake. IBP does work for this integral, but your earlier postings show that you do not have a grasp on how to apply it correctly.
     
  13. Jul 24, 2013 #12
    May be you are right. But I've just worked it using trigonometric substitution using x = tanθ and it worked correctly .. Maybe there exist a way to find it using IBP but it's a hard way and there is no use for it if there is an easier way :)
     
  14. Jul 24, 2013 #13

    LCKurtz

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    But since you are learning calculus, don't you think it might be of some value to figure out why it's not working for you?
     
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