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    ok just a few things i need help with.

    integration, denote to c

    3x^-1= 3in(x) + c or in(3x) + c ??

    x^-2= i can do this if it is x^-1 and x^-3, etc. just not 2, unless its 1/1x^-1 which surely cant be right?

    3/2 exp (3x) = how do you integrate if its a fraction? as in 3 over 2?

    erm check if i done this rght: 3 S (x^3 + x^2 +x) = 3 1/4x^4 + 1/3x^3 + x^2/2 + c ?? im worried about how you place the 3 into the answer have i done it right or do you need to do something else.

    Also could someone work through this example: that way i know how to use the quotient rule for differentiation!!

    x^2 + 2x + 4
    ___________
    x+2

    will appreciate the help people!!
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    Well, 3x^-1 means 3 . (x^-1), so it integrates to 3 ln x + c.

    The integral of x^-2 is (x^-1)/-1 = -x^-1 = -1/x.

    3/2 is just a constant. Treat it like you'd treat, say, the 4 in: 4 exp 3x.

    Your next integral - there should be a bracket round everything, so 3 (1/4x^4 + 1/3x^3 + x^2/2) + c. (Another acceptable answer would be 3 (1/4x^4 + 1/3x^3 + x^2/2 + c) - the 'c' is different in both cases, but that doesn't make any difference.)

    Quotient rule... \displaystyle \frac{\text{d}}{\text{d} x} \left( \frac{u}{v} \right) = \frac{v \frac{\text{d} u}{\text{d} x} - u \frac{\text{d}v}{\text{d} x}}{v^2} .

    So in your example:
    \displaystyle u = x^2+2x+4,\; v=x+2,\; \frac{\text{d} u}{\text{d} x} = 2x + 2,\; \frac{\text{d} v}{\text{d} x} = 1

    And so

    \displaystyle \frac{\text{d}}{\text{d} x} \left( \frac{x^2+2x+4}{x+2} \right) = \frac{(x+2)(2x+2) - (x^2+2x+2)(1)}{(x+2)^2} , which you can then simplify.

    Phew, long post, hope I didn't make any slips.
 
 
 
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Updated: August 27, 2007

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