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Simple Integral

  1. Jul 1, 2008 #1
    Hi

    I am trying to integrate

    [tex]x \sqrt{1+x^2}dx[/tex]

    by parts...but it seems to involve trigonometric functions - is it possible to solve this integral without using trig functions?


    Thx
     
  2. jcsd
  3. Jul 1, 2008 #2
    Why are you trying to use integration by parts? This is a very simple integral if you use an appropriate substitution.
     
  4. Jul 2, 2008 #3

    mathman

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    Gold Member

    The simple substitution is u2=1+x2.
     
  5. Jul 2, 2008 #4
    Mathman's solution is the easiest, but you probably could do it by parts.
     
  6. Jul 2, 2008 #5
    Might confuse him/her.

    Just let u = radican.

    u=1+x^2
     
  7. Jul 2, 2008 #6
    Hmm. Ok if u = 1 + x^2

    [tex] du = 2x dx [/tex]

    [tex]\int x \sqrt{1+x^2}dx = \frac{1}{2}\int \sqrt{1+x^2}2xdx[/tex]

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

    [tex]=\frac{1}{3} (1+x^2)^{\frac{3}{2}}[/tex]

    But if I take u^2=1+x^2

    [tex]u = \sqrt{1+x^2}[/tex]

    [tex]du = \frac{x}{\sqrt{1+x^2}}dx[/tex]

    [tex]\int{x\sqrt{1+x^2}dx} = \sqrt{1+x^2}\int{\sqrt{1+x^2}\frac{x}{\sqrt{1+x^2}}dx}[/tex]

    [tex]=\sqrt{1+x^2}\int{u du}[/tex]

    [tex]=\sqrt{1+x^2} \frac{{\sqrt{1+x^2}}^2}{2}[/tex]

    [tex]=\frac{(1+x^2)^{\frac{3}{2}}}{2}[/tex]

    I must be making a mistake somwhere...
     
  8. Jul 2, 2008 #7
    No, he meant to use u^2

    So ...

    [tex]u^2=1+x^2 \rightarrow udu=xdx[/tex]

    And remember that you can only pull out constants:

    [tex]c\int f(x)dx[/tex]
     
  9. Jul 2, 2008 #8
    wow, substitutions are so much mess. Try this way:

    x.sqrt(1+x^2)

    think about sqrt(1+x^2)
    it has integral
    (1+x^2)^3/2 --- [ref 1]

    it has derivative 3x (1+x^2)^0.5

    But, in original thing we only have x.sqrt(1+x^2)

    so, you need 1/3 with [ref 1]
     
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