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anonymous

  • 5 years ago

How do you find the integral of square root of 1+(1/x+1)^2 dx from x=0 to x=2?

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  1. anonymous
    • 5 years ago
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    HELLO

  2. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{2}\](1+(1/(X+1)^2))^1/2

  3. anonymous
    • 5 years ago
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    SIMPLIFY FIRST THE INSIDE

  4. anonymous
    • 5 years ago
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    1+ 1/(X+1)^2=((X+1)^2 +1)/(X+1)^2

  5. anonymous
    • 5 years ago
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    did u get it jin?

  6. anonymous
    • 5 years ago
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    this becomes \[\int\limits_{0}^{2}\](sqrt(x+1)^2 +1)dx/(x+1)

  7. anonymous
    • 5 years ago
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    now let this is S(u)^1/2 dx

  8. anonymous
    • 5 years ago
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    now from our simlified eq, let u=x+1, therefore S(u^1/2)dx=1/2 u]0 to 2,, we get:

  9. anonymous
    • 5 years ago
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    im sorry that is = 1/2 u^1/2 +2/2 =u^3/2

  10. anonymous
    • 5 years ago
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    whews,,sory again: =(u^3/2)/3/2 this is it

  11. anonymous
    • 5 years ago
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    =(2u^3/2)/3

  12. anonymous
    • 5 years ago
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    S u^3/2 du/2=

  13. anonymous
    • 5 years ago
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    =2S(1+(x+1)^2)^1/2dx/(2x+1)

  14. anonymous
    • 5 years ago
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    I will do this part for you \[\int\limits_{0}^{2} dx/(x+1)\] I assume this is what you meant \[\int\limits_{0}^{2} (dx+dx/(x+1)^{2})\]\[let u = (x+1) then du = dx\] let u = (x+1) then du = dx and \[\int\limits\limits dx/(x+1)^{2} =\int\limits\limits du/u ^{2} = u ^{-2+1}/(-2+1)\]=\[-1(1/(x+1)_{0}^{2}=2/3\] Now all you have to do is \[\int\limits_{0}^{2} dx\] Final answer is 8/3.

  15. anonymous
    • 5 years ago
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    \[\sqrt{1+u ^{2}} du/u\] I think you need to do a trig substitution for this, yes? let \[u=\sec \theta \] \[du=\sec^2\theta*d \theta\]

  16. anonymous
    • 5 years ago
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    question guys how do u connect the integral sign to dx \[\int\limits_{?}^{?}\]dx

  17. anonymous
    • 5 years ago
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    well sjin did u get it? that must be the way u solve it

  18. anonymous
    • 5 years ago
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    from my calculator the final ans is 2.301987535

  19. anonymous
    • 5 years ago
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    if you meant \[\int\limits du/(a ^{2} + u ^{2})^{n}\] where a =1 and u = (1/x+1) then you should look this up in a table of integrals

  20. anonymous
    • 5 years ago
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    if you meant \[\int\limits \sqrt({a}^{2}+u ^{2})du\] where a =1 and u = 1/(x+1) than again the best thing to to do is to look up the integral in a table of integrals or input the integral into a program like Maple. I hope I am not confusing you.

  21. anonymous
    • 5 years ago
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    thank you very much!

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