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Trouble with tables

  1. Oct 5, 2013 #1

    Jbreezy

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    1. The problem statement, all variables and given/known data

    Hi I have a broad quesiton. I'm trying to make something fit the form of something in one of the tables in my book I can't seem to quite make it.
    integral (1/2x^3 - 3x^2) dx


    2. Relevant equations



    3. The attempt at a solution

    I just messed with the denomator.
    1/(x^2(2x-3))

    I can't find something in the tables that fits this. Does anyone have any ideas?
     
    Jbreezy, Oct 5, 2013
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  3. Oct 5, 2013 #2

    Pranav-Arora

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    You can write the given integral as:
    $$\int \cfrac{dx}{x^2\cdot x\left(2-\cfrac{3}{x}\right)}$$
    A substitution would make it very easy. It is easy to spot.
     
    Pranav-Arora, Oct 5, 2013
  4. Oct 5, 2013 #3

    Jbreezy

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    I'm sorry I still don't see it. Sub for u = x^2?
     
    Jbreezy, Oct 5, 2013
  5. Oct 5, 2013 #4

    Pranav-Arora

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    You have x^2 in the denominator i.e 1/x^2. This is the derivative of something very familiar. :rolleyes:
     
    Pranav-Arora, Oct 5, 2013
  6. Oct 5, 2013 #5

    Mark44

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    Staff: Mentor

    What you wrote is this:
    $$ \int \frac 1 2 x^3 - 3x^2 dx$$

    To indicate that 2x3 - 3x2 is in the denominator, put parentheses around the denominator, not the whole fraction, like this 1/(2x3 - 3x2).
     
    Mark44, Oct 5, 2013
  7. Oct 5, 2013 #6

    Ray Vickson

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    Partial fractions.
     
    Ray Vickson, Oct 5, 2013
  8. Oct 5, 2013 #7

    Jbreezy

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    No sure. This would be -1/x whose derivavite is is 1/x^2 are you thinking ln of something? Ray says partial fractions but this is the section where you have to look them up in tables.
     
    Jbreezy, Oct 5, 2013
  9. Oct 5, 2013 #8

    Pranav-Arora

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    Yes, use the substitution 1/x=t.

    Partial fractions can also be used.
     
    Pranav-Arora, Oct 5, 2013
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