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Homework Help: Integral containing something over the root of x²-a²

  1. Jun 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate the next integrals, expressing it previously in the forms that contains [tex]\sqrt[ ]{a^2+u^2}[/tex], [tex]\sqrt[ ]{a^2-u^2}[/tex] and [tex]\sqrt[ ]{u^2-a^2}[/tex], and then solve it using the integral table;

    The exercise which I couldn't solve:

    [tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx[/tex]

    3. The attempt at a solution
    I started by completing the square:

    [tex]x^2+2x=(x+1)^2-1[/tex]

    Then

    [tex]u^2=(x+1)^2[/tex]
    [tex]u=(x+1)\Rightarrow{x+3=u+2}[/tex]
    [tex]du=dx[/tex]

    [tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx=\displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sqrt[ ]{u^2-1}}du[/tex]

    From here I've tried to solve it by parts
    [tex]t=u+2[/tex]
    [tex]dt=du[/tex]

    [tex]dv=\displaystyle\frac{1}{\sqrt[ ]{u^2-1}}du[/tex]
    [tex]v=\ln|u+\sqrt[ ]{u^2-1}|[/tex]

    [tex]\displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sqrt[ ]{u^2-1}}du=(u+2)\ln|u+\sqrt[ ]{u^2-1}|-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du[/tex]

    If I try again by parts with [tex]-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du[/tex] it gets more complicated.

    So, what do you say?

    Bye there.
     
  2. jcsd
  3. Jun 30, 2010 #2

    Dick

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    Homework Helper

    You can do integral u/sqrt(u^2-1) with a simple substitution. So split (u+2)/sqrt(u^2-1) up into two integrals.
     
  4. Jun 30, 2010 #3

    hunt_mat

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    Homework Helper

    The integral
    [tex]
    \int\frac{u+2}{\sqrt{u^{2}-1}}du
    [/tex]
    Can be computed by the substitution
    [tex]
    u=\cosh v
    [/tex]
    Everything should come out in the wash then
     
  5. Jun 30, 2010 #4
    Thank you both.

    This is what I wasn't seeing:
    Bye there!
     
  6. Jun 30, 2010 #5
    I've arrived to this solution:

    [tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C[/tex]
     
  7. Jun 30, 2010 #6

    Dick

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    Homework Helper

    Seems ok to me. I would write x^2+2*x instead of (x+1)^2-1. But that's up to you.
     
  8. Jun 30, 2010 #7
    Thanks Dick. BTW nice nickname :D
     
  9. Jul 1, 2010 #8
    yup, got the same answer !

    35bxok1.jpg

    P.S: anyone knows a program that writes LaTeX and has an intuitive interface? :)
     
  10. Jul 2, 2010 #9
  11. Jul 2, 2010 #10
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