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(x+1)^2 dx

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  1. Jul 8, 2015 #1
    Hi, I literally just registered so I have no idea about forum rules, also I'm not good in english.

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


    The equation is (x + 1)^2 dx.

    U = (x+1)
    DU = 1DX

    2. Relevant equations


    3. The attempt at a solution

    Here I get (U^3) over 3 times DX = (x^3 + 3x^2 + 3x + 1) over 3 times 1
    I simplify to (x^3) over 3 + x^2 + x +1 + C

    The answer shows the same as my attempt but without the + 1 at the end. What happened to + 1 ?
     
  2. jcsd
  3. Jul 8, 2015 #2
    First of all, it should be +1/3, not +1. Secondly, it gets absorbed into the C.

    Chet
     
  4. Jul 8, 2015 #3
    Ahhh, Thanks! I think I get it now, and you're right, it should be +1/3. So, basically whenever we have a constant we can just absorb it into the C? What if it was -1 instead of +1?
     
  5. Jul 9, 2015 #4

    Mark44

    Staff: Mentor

  6. Jul 9, 2015 #5

    Mentallic

    User Avatar
    Homework Helper

    The indefinite integral you're solving is basically asking the question, what is the function F such that
    [tex]\frac{d}{dx}F(x)=(x+1)^2[/tex]

    Now, [itex]\frac{d}{dx}(x^3/3+x^2+x)=(x+1)^2[/itex] so that is a valid answer, but it also turns out that we can add any constant (whether we know its value or not) and still end up with the same result after taking the derivative.

    Examples of valid constants

    [tex]+C[/tex]
    [tex]-C[/tex]
    [tex]+1+C[/tex]
    [tex]-1+C[/tex]
    [tex]+K+C[/tex]
    [tex]+C_1-C_2+\pi+\log_2{3}-10^{100}\sin{5}[/tex]

    I hope you get the idea. ANY number or group of numbers that are constants will essentially still be constant, and since it's customary to denote any arbitrary constant by C when calculating indefinite integrals, then you can merge whatever constants are left over in your result into the C. What is a little iffy however is if you end up with your result of 1/3+C and then decide to merge the 1/3 into C. It's allowed because C is a dummy constant that has no hidden value (unlike a variable when solving an equation), but your teacher might look down upon it. If you were ever disqualified in an exam for it though, you'd have a strong case to argue for the mark.
     
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