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Homework Help: Indefinite Integration by exchange of variables

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \int \frac {1}{\sqrt{x}(1+\sqrt{x})} dx [/tex]

    3. The attempt at a solution

    So let

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

    then

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

    So dx should be this:

    [tex] dx = 2x^{1/2} du[/tex]

    right?

    So now the Problem looks something like this:

    [tex] \int \frac {2\sqrt{x}}{\sqrt{x}(u)} du [/tex]

    [tex] \int \frac {\sqrt{x}}{(u)} du [/tex]

    I am stuck here, I am just not sure where to go from here, the answer is suppose to be:

    [tex] 2ln(1+\sqrt{x}) + C [/tex]

    but I really don't know how to get there.

    thanks!
     
  2. jcsd
  3. Apr 27, 2010 #2
    [tex] \int \frac {2\sqrt{x}}{u\sqrt{x}} du = \int \frac {\sqrt{x}}{u} du ??[/tex]

    [tex] \int \frac {2\sqrt{x}}{u\sqrt{x}} du = \int \frac {2}{u} du [/tex]
     
  4. Apr 27, 2010 #3
    haha right.... so then the proper way to get to the answer is this then:

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

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

    then take an antiderivative:

    [tex] 2*ln(u) [/tex]

    [tex] 2ln(1+\sqrt{x}) +C[/tex]

    is that right?
     
  5. Apr 27, 2010 #4
    Looks right to me.
     
  6. Apr 27, 2010 #5
    ok can you help me on a problem of the same type?

    [tex] \int \frac {3t+4}{5-t}dt [/tex]

    Let:

    [tex] u = 5 -t [/tex]

    [tex] du = -1 dt [/tex]

    [tex] dt = -1 du [/tex]

    [tex] \int -\frac {3t+4}{u} du [/tex]

    so here I am kind of guessing..

    [tex] \int -(3t+4)* \frac {1}{u} du [/tex]

    [tex] \int (-3t-4) * \frac {1}{u}du [/tex]

    [tex] \int -3t\frac{1}{u}-4\frac{1}{u} [/tex]

    I don't think I am on the right track here though, the correct answer is

    [tex] -3t -19ln(5-t) +C [/tex]

    and I don't see how to get there from here, its probably just something I am not seeing again... thanks in advance
     
  7. Apr 27, 2010 #6
    [tex] u = 5 -t [/tex] so
    [tex] t = 5 -u [/tex]
    [tex] 3t + 4 = 3(5-t)+4 = 19 -3u [/tex]

    Try to figure it out from there. Hint: split into 2 integrals. You have basically right idea, you just need to get that t in terms of u somehow so you can actually integrate.
     
  8. Apr 27, 2010 #7
    ok so starting from the point of

    [tex] \int (-19+3u)*\frac{1}{u} du [/tex]

    note that I just applied the - sign throughout, thats why my signs are different than yours

    so my next guess would be here:

    [tex] \int -19\frac{1}{u} + 3u\frac{1}{u} du [/tex]

    so then I am assuming I can like so:

    [tex] \int -19\frac{1}{u} + 3u\frac{1}{u} du = \int -19\frac{1}{u} du + \int \frac {3u}{u} du [/tex]

    so I can see that the -19ln(5-t) comes out there but the integral of 3 du is 3u right? so 3(5-t) = 15-3t which does give me the -3t for the solution but the +15 ruins the answer... at least I think so...
     
  9. Apr 27, 2010 #8
    I got the same answer as you and that looks right. The answer you got it:

    [tex] y = 15 - 3t -19ln(5-t) + c_1 [/tex]

    Let

    [tex] c = 15 + c_1 \rightarrow y = -3t - 19ln(5-t) + c [/tex]

    The +15 is just a constant so it does not change the functional form of your answer. That is why you might as well absorb it into the [tex]c_1[/tex] so your answer looks nicer.
     
  10. Apr 27, 2010 #9
    Wow I actually didn't know that, thanks man, I would have been thinking that over forever.
     
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