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Integration by substitution

  1. Jun 1, 2016 #1
    1. The problem statement, all variables and given/known data
    dz/dx=(3x2+x)(2x^3+x^2)^2



    2. Relevant equations

    ∫(3x^2+x)(2x^3+x^2)^2 dx
    3. The attempt at a solution
    I tried substituting (2x^3+x^2)
    Let t= 2x^3 + x^2
    dt=6x^2+2x dx
    dt/dx= 6x^2+2x
    I can only solve till this point . I don't have any clue how to solve it further
    But how do we get 1/2 ∫u^2 du ?? I don't understand it at all
     
  2. jcsd
  3. Jun 1, 2016 #2

    Math_QED

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    Let u = 2x^3 + x^2
    => du = (6x^2 + 2x)dx
    => du = 2(3x^2 + x)dx
    => du/2 = (3x^2 + x)dx

    Thus, the integral becomes what you need.
     
  4. Jun 1, 2016 #3

    SammyS

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    Well, you used t as a variable rather than u, but that's unimportant.

    Looking at Math_QED's post, the main thing your solution is lacking is that you should factor a 2 out of the right hand side of ##\ dt=(6x^2+2x) dx \ .##
     
  5. Jun 3, 2016 #4
    Thank you both . I think i understand it now :)
     
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