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Integration problems

  1. Apr 9, 2005 #1

    mad

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    Hello everyone.
    I have been trying to do those 2 exercices for a while now and I can't get it.. We just started doing Integration by parts (is that how you call it in english?)
    here are the problems

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


    for the 2nd one I tried
    u=e^(x^2) dv= x^3
    du= 2xe^(x^2) v= (x^4)/4

    From there I just get stuck..
    For #2, I dont even know how to start it because there are 3 terms.. Can you do that by parts?
    I know these are rookies question, but please explain me
    thanks :)
     
    Last edited: Apr 9, 2005
  2. jcsd
  3. Apr 9, 2005 #2

    quasar987

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    1) Try your luck with u = x² and dv = xe^x² (you can find v by substitution with x² = w)
     
  4. Apr 9, 2005 #3

    quasar987

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    2) Set

    [tex]u = (x+1)^2 \ \ \ dv = xe^x dx \Rightarrow du = (2x^2 + 4x +2)dx \ \ \ v = \int(xe^x)dx[/tex]

    Let us find v by integrating by parts, using

    [tex]w = x \ \ \ dz = e^x dx \Rightarrow dw = dx \ \ \ z = e^x \Rightarrow v = xe^x - \int e^x dx = xe^x - e^x [/tex]

    [tex]v = xe^x - e^x[/tex]

    Continue. You will have to do many other integrations by parts to get to the final answer.
     
  5. Apr 9, 2005 #4

    dextercioby

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    The last one is tricky.

    [tex] \int\frac{x}{(x+1)^{2}} \ dx=\int\frac{(x+1)-1}{(x+1)^{2}} \ dx=\int \frac{dx}{x+1} -\int \frac{d(x+1)}{(x+1)^{2}} =\ln (x+1)+\frac{1}{x+1} [/tex]

    Now

    [tex] I:=\int e^{x}\frac{x}{(x+1)^{2}} \ dx [/tex]

    can be part integrated to get

    [tex] I=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)-\int e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right) \ dx=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)[/tex]

    [tex] -\int e^{x} \ln (x+1) \ dx-\int \frac{e^{x}}{x+1} \ dx [/tex]

    Use part integration for the first of the last 2 integrals

    [tex] I=e^{x}\left(\ln (x+1)+\frac{1}{x+1}\right)-\left[e^{x}\ln(x+1)\right]+\int \frac{e^{x}}{x+1} \ dx-\int \frac{e^{x}}{x+1} \ dx =\frac{e^{x}}{x+1} + C [/tex]

    Daniel.
     
    Last edited: Apr 9, 2005
  6. Apr 10, 2005 #5

    quasar987

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    Dexter: What does "Docendo discitur" mean? Who's Seneca?
     
  7. Apr 10, 2005 #6

    dextercioby

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    I'll let u discover who Seneca was.U can learn more from the internet.

    "You learn by teaching others".It should be a motto for these forums...

    Daniel.
     
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