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What is the antiderivative of x^2 + (5/x^5) +cos9x - e^-7x?

  1. Dec 2, 2007 #1
    What is the antiderivative of x^2 + (5/x^5) +cos9x - e^-7x? I'm so lost...

    and also (2x + 5sqrtx)/ x^2
     
  2. jcsd
  3. Dec 2, 2007 #2

    arildno

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    An anti-derivative of a sum is a sum of anti-derivatives.
     
  4. Dec 3, 2007 #3
    Well, do you know basic integration rules for powers, exponential, and trigonometric functions?
     
  5. Dec 4, 2007 #4

    HallsofIvy

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    Do it one step at a time (that's what the others are saying). What is the anti-derivative of x2? What is the anti-derivative of 5x-5? What is the anti-derivative of cos(9x)? What is the anti-derivative of e-7x? If you don't know those, you should go back and review your basic formulas.

    As for [tex]\frac{2x+ 5\sqrt{x}}{x^2}[/tex], that's really the same as
    [tex]\frac{2x}{x^2}+ \frac{5x^{\frac{1}{2}}}{x^2}= 2x^{-1}+ 5x^{-\frac{3}{2}[/tex].
    The anti-derivatives of those should be easy- use the first anti-derivative formula you learned!
     
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