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Integration problem

  1. Jan 21, 2014 #1
    Hello im trying to integrate this problem:
    ∫x-1/x^2-2x-3







    My attempt solution is differ x^2 - 2x + 4 = 2x - 2 and basically the top x-1 can be manipulate by multiplying 2 this would give me the answer as 2 ln|x^2-2x+4| + c im not sure im right can somebody help me out? thank you!
     
  2. jcsd
  3. Jan 21, 2014 #2
    You are almost correct. This is where you went wrong, you should recognize this as a simple u-substitution. So you would set u=x^2-2x-3 and du=2x-2 dx. And you were correct in recognizing that it only differs the numerator by a constant, but what you would want to do is now divide du=2x-2 by 2. This gives (1/2)du=x-1dx. Now you can make the substitution. And you get ##(1/2)[integral]du/u= (1/2)ln(u)+c## then just plug back in for u and there's your answer.

    Just try to physically write down the u substitution and everything that follows and you won't lose coefficients like that.
     
  4. Jan 21, 2014 #3
    thanks !! sorry and for this one ∫x-3 / x^2+2x-3 .. i think cant do the same as the previous one because would be bottom 2x + 2 but in the top is x - 3 so is a bit impossible to do the previous method. but can i do by partial fractions (x+3) (x-1)
     
  5. Jan 21, 2014 #4

    Ray Vickson

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    You should have no trouble computing
    [tex] \int \left( x - \frac{3}{x^2} + 2x - 3 \right) \, dx[/tex]
    which is what you wrote. If you meant something else, you must use parentheses.
     
  6. Jan 21, 2014 #5
    This one is the two parts, the first part will be the exact same as the previous integral. The next part will be just as easy. How can you break this into two integrals one of which is exactly the same as the previous problem?
     
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