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How to determine this integral? Thank you

  1. Dec 12, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-12-12_17-11-48.png

    2. Relevant equations
    k∫[ƒ(x)]n ƒ'(x) dx

    3. The attempt at a solution
    i tried to using algebraic substitution to determine that i had let u = 1-x or X2-2x+1 or x or root(x) but it still cannot solve it.
    Please give me hint how to solve it.
    Thank you
     
  2. jcsd
  3. Dec 12, 2016 #2

    Jonathan Scott

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    I'd just expand it as a sum of the three terms, each of which is a half-integer power of x, integrate each separately and then simplify the result sum.
     
  4. Dec 12, 2016 #3
    Expand ##(1-x)^2=1-2x+x^2## and then treat each term separately.
     
  5. Dec 12, 2016 #4

    PeroK

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    The substitution ##u = \sqrt{x}## should have simplified it. Perhaps you ought to post what you did to see where you went wrong.
     
  6. Dec 12, 2016 #5
    $$ ∫ \frac{ (1-x)^2 }{ 2 \sqrt{x} } dx$$
    $$ =∫ \frac{ (x^2+2x+1) }{ 2 \sqrt{x} } dx$$
    $$ Let u = \sqrt{x} $$
    $$ dx= \frac{ 2du }{ x^{\frac{ -1 }{ 2 }} } $$
    $$ ∫\frac{ (x^2+2x+1) }{ 2u } \frac{ 2du }{ x^{\frac{ -1 }{ 2 }} }$$

    and then i have no idea what is next step
    thank you very much
     
  7. Dec 12, 2016 #6

    PeroK

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    The point of a substitution is to replace one variable with another, not to mix the two. You need to replace all the terms in ##x## with the relevant term in ##u##.
     
  8. Dec 12, 2016 #7

    Ray Vickson

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    Using any kind of substitution in this problem is unhelpful; it just makes the problem harder, not easier! Just integrate the three terms separately.
     
  9. Dec 12, 2016 #8
    thank you ~:smile:
    The answer:
    $$ \frac{ x \frac{ 5 }{ 2 } }{ 5 } + \frac{ 2x \frac{ 3 }{ 2 } }{ 3 } + \sqrt{x} $$
    Am i right ?
     
  10. Dec 12, 2016 #9

    PeroK

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    If the OP doesn't understand how to do substitution, it's as well to get that fixed. He'll need to get to grips with it sooner or later!
     
  11. Dec 12, 2016 #10

    Jonathan Scott

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    You lost a minus sign when you expanded the square, and your formatting is a bit scrambled.
     
  12. Dec 12, 2016 #11

    PeroK

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    You should never (or rarely) have to ask whether an indefinite integral answer is correct. You should differentiate your answer and see whether you get the original function back. You have, however, forgotten the constant of integration.
     
  13. Dec 12, 2016 #12
    ok ~ thank you so much
     
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