## 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?

1. anonymous

HELLO

2. anonymous

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

3. anonymous

SIMPLIFY FIRST THE INSIDE

4. anonymous

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

5. anonymous

did u get it jin?

6. anonymous

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

7. anonymous

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

8. anonymous

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

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

10. anonymous

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

11. anonymous

=(2u^3/2)/3

12. anonymous

S u^3/2 du/2=

13. anonymous

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

14. 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.

15. anonymous

$\sqrt{1+u ^{2}} du/u$ I think you need to do a trig substitution for this, yes? let $u=\sec \theta$ $du=\sec^2\theta*d \theta$

16. anonymous

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

17. anonymous

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

18. anonymous

from my calculator the final ans is 2.301987535

19. 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

20. 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.

21. anonymous

thank you very much!