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Integral (x*e^(-3x)). How the !

  1. Jun 10, 2007 #1
    Hello, Forum!

    I just registered after seeing you actually help people understand their problems. That's great.

    We have (or should have) learned about linearity, substitution and partial integration. However, I don't know when to use which! Could someone also give me a bit of an expanation on this? :(

    I have to solve an integral:
    x*e^(-3x) dx

    My train of thought: I have almost got 2 'basis integrals': x dx and e^x dx. I probably need to substitute to get them to the basic form. But how!
    As you see I'm pretty clueless, but what I came up with was:
    u = -3x --> u'= -3
    v' = x --> v = (x²)/2
    However, this leads nowhere. I don't know what to do!

    According to derive, the solution is supposed to be:
    Code (Text):

         1           -3x  ⎛    x            1       ⎞
    ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ - e           ⎜⎯⎯⎯⎯⎯⎯⎯⎯⎯ + ⎯⎯⎯⎯⎯⎯ ⎟
              2            ⎜ 3·LN(e)              2⎟
     9·LN(e)             ⎝               9·LN(e) ⎠
    I sincerely hope someone will be able to show me the light!
    Thanks in advance.

    PS: Our teacher is really bad at teaching!
    Last edited: Jun 10, 2007
  2. jcsd
  3. Jun 10, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Try u = x and v' = e^(-3x)

    In integration by parts what you're looking to do is reduce any x factors to a constant and thus you set those equal to u. This reduces the right hand side integral to a single function which should be easy to deal with.

    If the x factors are higher powers then apply the integration by parts method until the x reduces to a constant or you can come up with a reduction formula.

    The substitution method is a little intuitive because you're looking for something that is a derivative of something else in the function. Just keep practising some substitution questions and you will soon start to spot them fairly easily.

    For example:

    [tex] \int xe^{x^2} dx [/tex]

    You can spot that x is almost the derivative of x2. So we use the following substitution:

    [tex] u=x^2 [/tex] therefore [tex]\frac{du}{dx}=2x \Rightarrow xdx=\frac{1}{2}du [/tex]

    [tex] \frac{1}{2} \int e^u du = \frac{1}{2} e^{x^2} + c [/tex]
    Last edited: Jun 10, 2007
  4. Jun 10, 2007 #3
    The LIATE rule helps you identify which one to use as u.
    Order of priority is:

    Logarithms, Inverse Trigonometric, Algebraic, Trigonometric, Exponential
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