anonymous
  • anonymous
How do you find the integral of square root of 1+(1/x+1)^2 dx from x=0 to x=2?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
HELLO
anonymous
  • anonymous
\[\int\limits_{0}^{2}\](1+(1/(X+1)^2))^1/2
anonymous
  • anonymous
SIMPLIFY FIRST THE INSIDE

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anonymous
  • anonymous
1+ 1/(X+1)^2=((X+1)^2 +1)/(X+1)^2
anonymous
  • anonymous
did u get it jin?
anonymous
  • anonymous
this becomes \[\int\limits_{0}^{2}\](sqrt(x+1)^2 +1)dx/(x+1)
anonymous
  • anonymous
now let this is S(u)^1/2 dx
anonymous
  • anonymous
now from our simlified eq, let u=x+1, therefore S(u^1/2)dx=1/2 u]0 to 2,, we get:
anonymous
  • anonymous
im sorry that is = 1/2 u^1/2 +2/2 =u^3/2
anonymous
  • anonymous
whews,,sory again: =(u^3/2)/3/2 this is it
anonymous
  • anonymous
=(2u^3/2)/3
anonymous
  • anonymous
S u^3/2 du/2=
anonymous
  • anonymous
=2S(1+(x+1)^2)^1/2dx/(2x+1)
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\sqrt{1+u ^{2}} du/u\] I think you need to do a trig substitution for this, yes? let \[u=\sec \theta \] \[du=\sec^2\theta*d \theta\]
anonymous
  • anonymous
question guys how do u connect the integral sign to dx \[\int\limits_{?}^{?}\]dx
anonymous
  • anonymous
well sjin did u get it? that must be the way u solve it
anonymous
  • anonymous
from my calculator the final ans is 2.301987535
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
thank you very much!

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