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Integration U substitution then square it I think.

  1. Mar 31, 2013 #1
    1. ∫ 1/(2√(x+3)+x)



    2. Not sure if I'm beginging this correctly or not but I get stuck.



    3. Let u= √x+3 then u2 = x+3 2udu=dx dx=2√[(x+3)

    Therefore: ∫1/u2-3 Not sure where to go from here?
     
  2. jcsd
  3. Mar 31, 2013 #2
    Ok what you did is good but take x from your u^2 you will get x=u^2-3. Then substitute it back into you integral and it should be simple from there.
     
  4. Mar 31, 2013 #3
    So plug it into the original integral and get ∫1/(2√(u2-3+3) ????
    So the 3's equal 0

    Gives you: ∫1(2√u2)

    Sqrt and 2 cancel leaving you 1/2u

    but then that gives you 2u-1

    integrating that gives you u0 which is where I keep getting stuck.
     
  5. Mar 31, 2013 #4
    If I sub the other way and put x in for u^2-3 then that gives me 1/x. Integral of that is ln absvalue x. When I plug in numbers to make it a definite integral and check it on my calc. that doesn't work.
     
  6. Mar 31, 2013 #5

    Curious3141

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    I didn't read the rest of this, but this is wrong. ##\int \frac{1}{u}du = \ln |u| + c## (c is the constant).

    The rule ##\int u^n du = \frac{1}{n+1}u^{n+1} + c## only applies for ##n \neq -1##.
     
  7. Mar 31, 2013 #6
    So you can pull out the 1/2 and make it .5∫(1/u)

    The integral of that is .5(ln absvalue(u))

    Sub u back in and get

    .5(ln(√x+3)) abs value of √x+3 of course

    that's still not giving me a right answer though when I check it.
     
  8. Mar 31, 2013 #7

    SammyS

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    No, you don't get 2u-1 .

    [itex]\displaystyle \frac{1}{2\sqrt{x+3}+x}\ [/itex] becomes [itex]\displaystyle \ \frac{1}{2u+u^2-3}\ .[/itex]
     
  9. Mar 31, 2013 #8
    -1/u+ln(2u)-u/3 Just plug in √x+3 back in for u.


    So that gives you -(1/√x+3)+ln(2√x+3)-(√x+3)/3) ?
     
  10. Mar 31, 2013 #9

    SammyS

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    It's not clear how you got that.

    The integral [itex]\displaystyle \ \int \frac{dx}{2\sqrt{x+3}+x}\ [/itex] becomes [itex]\displaystyle \ \int \frac{2u\,du}{u^2+2u-3}\ .[/itex]

    Now use partial fractions.
     
  11. Mar 31, 2013 #10
    You mean partial fraction decomposition or just split them up into separate fractions?
     
  12. Mar 31, 2013 #11

    SammyS

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    Yes, partial fraction decomposition. How else can you split them up into separate fractions?
     
  13. Mar 31, 2013 #12
    Can you make the u2+2u-3 = (u+3)(u-1)

    Then A/(u+3) + B/(u-1) ?
     
  14. Mar 31, 2013 #13

    SammyS

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    Try it !
     
  15. Mar 31, 2013 #14
    So 2=A(u-1)+B(u+3)

    Let u= 1

    2=A(0)+B(4)

    2=3B so B=(1/2)

    Let u=-3 and A=-1/2

    When you put those in can you just bring the 1/2's out in front of the fraction or do you have to distribute then through the bottom of the integral?
     
  16. Mar 31, 2013 #15
    -1/2∫1/(u+3)+1/2∫1/(u-1)

    ends up being 1/2* ln(√x+3)-1/√x+3)+3
     
  17. Mar 31, 2013 #16

    SammyS

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    Try to be more careful.

    That should be:

    2u = A(u-1) + B(u+3) .
     
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