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Homework Help: Integration, where have i gone wrong?

  1. Feb 6, 2009 #1
    [tex]\int[/tex]arctg[tex]\sqrt{x}[/tex]dx

    using[tex]\int[/tex]udv=uv-[tex]\int[/tex]vdu

    dv=dx
    v=x

    u=arctg[tex]\sqrt{x}[/tex]
    du=u'dx=[1/(1+x)]dx

    [tex]\int[/tex]arctg[tex]\sqrt{x}[/tex]dx=xarctg[tex]\sqrt{x}[/tex] - [tex]\int[/tex]x*[1/(1+x)]dx
    =xarctg[tex]\sqrt{x}[/tex] - [tex]\int[/tex][x/(1+x)]dx
    =xarctg[tex]\sqrt{x}[/tex] - [tex]\int[/tex]1-[1/(1+x)]dx
    =xarctg[tex]\sqrt{x}[/tex] - x+ln|x+1|+c
    but this is wrong, where have i made the mistake?
     
  2. jcsd
  3. Feb 6, 2009 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    The derivative of arctan(sqrt(x)) is (1/(1+x))*1/2(sqrt(x)), you can still solve this question by looking at:
    sqrt(x)/(1+x)dx=2/3(d(x^3/2))/(1+(x^3/2)^(2/3))
    sqrt(x)dx=2/3d(x^3/2)
    and the integral of dy/(1+y^2/3)
    can be solved by writing y^1/3=sinh(t).

    I am not sure this will work, but I think that there is no easy way here.
     
  4. Feb 6, 2009 #3
    thanks
     
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