Question

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

Answer 1

HELLO

Answer 2

\[\int\limits_{0}^{2}\](1+(1/(X+1)^2))^1/2

Answer 3

SIMPLIFY FIRST THE INSIDE

Answer 4

1+ 1/(X+1)^2=((X+1)^2 +1)/(X+1)^2

Answer 5

did u get it jin?

Answer 6

this becomes
\[\int\limits_{0}^{2}\](sqrt(x+1)^2 +1)dx/(x+1)

Answer 7

now let this is S(u)^1/2 dx

Answer 8

now from our simlified eq, let u=x+1,
therefore S(u^1/2)dx=1/2 u]0 to 2,, we get:

Answer 9

im sorry that is = 1/2 u^1/2 +2/2 =u^3/2

Answer 10

whews,,sory again:
=(u^3/2)/3/2 this is it

Answer 11

=(2u^3/2)/3

Answer 12

S u^3/2 du/2=

Answer 13

=2S(1+(x+1)^2)^1/2dx/(2x+1)

Answer 14

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.

Answer 15

\[\sqrt{1+u ^{2}} du/u\]

I think you need to do a trig substitution for this, yes?

let \[u=\sec \theta \]
\[du=\sec^2\theta*d \theta\]

Answer 16

question guys how do u connect the integral sign to dx \[\int\limits_{?}^{?}\]dx

Answer 17

well sjin did u get it? that must be the way u solve it

Answer 18

from my calculator the final ans is 2.301987535

Answer 19

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

Answer 20

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.

Answer 21

thank you very much!