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Solve ∫(x+2)√(x-3)dx

  1. Oct 1, 2005 #1
    Hoping anyone could give a hand on this....

    this is what i've gotten so far...

    ∫(x+2)√(x-3)dx
    u=(x-3)^1/2
    u^2=(x-3)
    x=u^2-3
    dx=2u du

    ∫(x+2)√(x-3)dx=(u^2-3+2)*(u)*(2u) du =2∫(u^2-1)*(u)*(u)
    = 2∫(u^2-1)*(u^2)
    =2∫(u^4-u^2)
    =(2*1/5*u^5) - ((2/5)*2*(1/3)*u^3)
    =2/5*(x-3)^(5/2) - 4/15(x-3)^3/2 + K

    Is it right?

    Thank you so much guys...
     
  2. jcsd
  3. Oct 1, 2005 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    It would be much easier to write x + 2 as x - 3 + 5 and proceed from there.
     
  4. Oct 2, 2005 #3

    hotvette

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    Homework Helper

    One sure way of checking yourself is to differentiate the result and see if you come with the original integrand.
     
  5. Oct 2, 2005 #4
    ∫(x+2)√(x-3)dx
    ∫(x-3+5)√(x-3)dx

    u=(x-3)^1/2
    u^2=(x-3)
    x=u^2-3
    dx=1/du

    ∫(x+2)√(x-3)dx=(u^2+5)*(u^2) du
    = 5∫(u^4) du +∫(u^2)
    =5/5*u^5 + 1/3*u^3
    =(x-3)^(5/2) - 1/3(x-3)^3/2 + K

    Is it right now?

    Thank you very much...


    Thank you so much guys...
     
  6. Oct 2, 2005 #5
    I think the last one is wrong... here is the one I think might be right...
    Can you please tell me if it is, and if not I would really appreciate if you could show me the way... thanks....


    ∫(x+2)√(x-3)dx
    ∫(x-3+5)√(x-3)dx

    u=(x-3)^1/2
    u^2=(x-3)
    x=u^2-3
    dx=2u

    ∫(x+2)√(x-3)dx=(u^2+5)*(u)*(2u) du
    = 2∫(u^4+ 5u^2)
    = 2∫(u^4 du +5∫u^2)
    = 2*1/5*d^5 + 2*5*1/3*u^3
    = 2/5*u^5 + 10/3*u^3
    = 2/5(x-3)^(3/2) + 10/3(x-3)^(3/2) + K


    Is it right now?

    Thank you very much...


    Thank you so much guys...
     
  7. Oct 2, 2005 #6
    I think I finally got this....


    ∫(x+2)√(x-3)dx
    ∫(x-3+5)√(x-3)dx

    u=(x-3)^1/2
    u^2=(x-3)
    x=u^2-3
    dx=2u

    ∫(x+2)√(x-3)dx=(u^2+5)*(u)*(2u) du
    = 2∫(u^4+ 5u^2)
    = 2∫(u^4 du +5∫u^2)
    = 2*1/5*u^5 + 2*5*1/3*u^3
    = 2/5*u^5 + 10/3*u^3
    u*3(2/5u^2 +10/3) + k
    (x-3)^(3/2) * (2/5(x-3) + 10/3) + K
    (x-3)^(3/2) * (2/5x - 6/5 + 10/3) + k
    (x-3)^(3/2) * (2/5x - 18/15 + 50/3) + k
    (x-3)^(3/2) * (2/5x + 32/15) + k

    Is it right now?

    Thank you so much guys...
     
  8. Oct 2, 2005 #7

    VietDao29

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    Homework Helper

    It looks correct. But you seem to make the problem more complicated. Here's a shorter one:
    [tex]\int (x + 2) \sqrt{x - 3} dx = \int (x - 3 + 5) \sqrt{x - 3} dx = \int \left( (x - 3) ^ {\frac{3}{2}} + 5 (x - 3) ^ {\frac{1}{2}} \right) dx[/tex]
    [tex]\int (x - 3) ^ {\frac{3}{2}} dx + \ \int 5 (x - 3) ^ {\frac{1}{2}} dx = \int (x - 3) ^ {\frac{3}{2}} d(x - 3) + \ 5 \int (x - 3) ^ {\frac{1}{2}} d(x - 3)[/tex]
    [tex]= \frac{2}{5}(x - 3) ^ {\frac{5}{2}} + \frac{10}{3} (x - 3) ^ {\frac{3}{2}} + C[/tex]
    -------------
    Or you can just stop here and substitute u in:
    [tex]u ^ 2 = x - 3 \Rightarrow u = \sqrt{x - 3}[/tex]
    [tex]u ^ 5 = (x - 3) ^ {\frac{5}{2}}[/tex]
    [tex]u ^ 3 = (x - 3) ^ {\frac{3}{2}}[/tex]
    Viet Dao,
     
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