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Homework Help: Best method to solve this integral?

  1. Mar 26, 2006 #1

    G01

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    [tex]\int\frac{dx}{x^2\sqrt{4x+1}} [/tex] Can someone give me a hint? I've been working on this forever.
     
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  3. Mar 26, 2006 #2

    benorin

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    Let [tex]u=\sqrt{4x+1} \Rightarrow du=\frac{4dx}{\sqrt{4x+1}}[/tex]
    also, [tex]\frac{1}{x^2}=\frac{16}{(u^2-1)^2}[/tex]
    then use trig substitution
     
  4. Mar 26, 2006 #3
    NVM, my calculation was wrong. but i can tell you that the answer involves arctanh.
     
    Last edited: Mar 26, 2006
  5. Mar 26, 2006 #4

    benorin

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    This will give

    [tex]\int\frac{dx}{x^2\sqrt{4x+1}} = 4\int \frac{du}{(u^2-1)^2}[/tex]
     
  6. Mar 26, 2006 #5

    G01

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    ok using trig substitution, i got the integral of cos^2x / sin^3x dx. Is this correct so far?
     
  7. Mar 26, 2006 #6

    G01

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    OK i think i may have gotten this. Is this the answer.

    [tex] \frac{-\sqrt{4x + 1}}{2x} - 2\ln|\frac{\sqrt{4x + 1} - 1}{\sqrt{4x}}| +C [/tex]
     
  8. Mar 26, 2006 #7

    benorin

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    I get

    [tex]\int\frac{dx}{x^2\sqrt{4x+1}} = \log \left| \frac{\sqrt{4x+1}+1}{\sqrt{4x+1}-1}\right| -\frac{\sqrt{4x+1}}{2x}+C[/tex]

    you can always differentiate to check.

    EDIT: I forgot the factor of 4.
     
    Last edited: Mar 26, 2006
  9. Mar 26, 2006 #8

    benorin

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    EDIT: Oops! should be:

    Let [tex]u=\sqrt{4x+1} \Rightarrow du=\frac{1}{2}\frac{4dx}{\sqrt{4x+1}}[/tex]
    also, [tex]\frac{1}{x^2}=\frac{16}{(u^2-1)^2}[/tex]

    to give

    [tex]\int\frac{dx}{x^2\sqrt{4x+1}} = 8\int \frac{du}{(u^2-1)^2}= 2\log \left| \frac{\sqrt{4x+1}+1}{\sqrt{4x+1}-1}\right| -\frac{\sqrt{4x+1}}{x}+C[/tex]
     
    Last edited: Mar 26, 2006
  10. Mar 26, 2006 #9

    benorin

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    BTW, for real x,

    [tex]\mbox{arctanh}^{-1}(x) = \frac{1}{2}\log\left| \frac{x+1}{x-1}\right|[/tex]

    which would help if you went to www.integrals.com
     
  11. Mar 27, 2006 #10

    G01

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    [tex]8\int \frac{du}{(u^2-1)^2}[/tex]

    Im screwing up this integral somwhere. Did you use u = sec[tex]\theta[/tex]? And did you end up with:

    [tex]-4\csc\theta\cot\theta - 4\ln|\csc\theta - \cot\theta| + C [/tex]

    Thanks alot for all of this help.
     
  12. Mar 27, 2006 #11

    benorin

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    You didn't screw-up, I did. My bad, do this rather:

    [tex]\frac{8}{(u^2-1)^2}=\frac{8}{(u-1)^2(u+1)^2}[/tex]

    and now partial fractions

    [tex]\frac{8}{(u-1)^2(u+1)^2}=\frac{A}{u-1}+\frac{B}{(u-1)^2}+\frac{C}{u+1}+\frac{D}{(u+1)^2}[/tex]

    cross-multiply to get

    [tex]8=A(u-1)(u+1)^2+B(u+1)^2+C(u-1)^2(u+1)+D(u-1)^2[/tex]

    plug-in u=1 to get 8=4B or B=2;

    plug-in u=-1 to get 8=4D or D=2;

    plug-in u=0 to get 8=-A+B+C+D, but B=D=2, so 4=-A+C

    plug-in u=2 to get 8=9A+9B+3C+D, but B=D=2, so -12=9A+3C
    solving these two equations gives A=-2 and C=2. Finally, we get

    [tex]\int\frac{8du}{(u-1)^2(u+1)^2}=\int\left(\frac{-2}{u-1}+\frac{2}{(u-1)^2}+\frac{2}{u+1}+\frac{2}{(u+1)^2}\right) du = -2\log |u-1|-\frac{2}{u-1}+2\log |u+1|-\frac{2}{u+1}+C[/tex]
    [tex]= 2\log \left| \frac{u+1}{u-1}\right|-\frac{4u}{u^2-1}+C[/tex]

    but [tex]u=\sqrt{4x+1}[/tex] so...
     
  13. Mar 27, 2006 #12

    G01

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    ok thanks alot man....
     
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