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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2}\](1+(1/(X+1)^2))^1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0SIMPLIFY FIRST THE INSIDE

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01+ 1/(X+1)^2=((X+1)^2 +1)/(X+1)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this becomes \[\int\limits_{0}^{2}\](sqrt(x+1)^2 +1)dx/(x+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now let this is S(u)^1/2 dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now from our simlified eq, let u=x+1, therefore S(u^1/2)dx=1/2 u]0 to 2,, we get:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorry that is = 1/2 u^1/2 +2/2 =u^3/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whews,,sory again: =(u^3/2)/3/2 this is it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0=2S(1+(x+1)^2)^1/2dx/(2x+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{1+u ^{2}} du/u\] I think you need to do a trig substitution for this, yes? let \[u=\sec \theta \] \[du=\sec^2\theta*d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0question guys how do u connect the integral sign to dx \[\int\limits_{?}^{?}\]dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well sjin did u get it? that must be the way u solve it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from my calculator the final ans is 2.301987535

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0if 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.
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