## anonymous 5 years ago what is the intergral of 1/1 +x^2 dy

1. anonymous

I think it is( x^3)/(3)+c

2. amistre64

1/1 = 1

3. amistre64

but that would be to obvious to be right :)

4. amistre64

and dy? instead of dx? is that a typo....

5. amistre64

(S) 1/(1+x^2) dx perhaps...

6. amistre64

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

7. amistre64

ln(1+x^2) --------- (2x) I wish I knew if I am doing this right :)

8. anonymous

Yes, it's ∫1/(1+x^2)dx. Turns out to be arctan(x), I believe...it's on the tables too.

9. amistre64

DOH!!.... yeah.....

10. amistre64

and I messed p taking that "x" out of it.... should left it in if I was going to do what I was thinking of doing :)

11. amistre64

arctan(x)... 1/sqrt(x^2+1)?? great, i just confused myself lol

12. anonymous

let consider u = 1+x^2 then du = 2x dx and dx = 1/2x du substitute to the integral S u/1/2x du =1/2x.1/2(1+x^2)^2 =1/4x ( 1+ X^2 )^2 would be the answer....

13. anonymous

quantum that answer is correct but how did you work it? bysetting it up 1/2 tan^-1 +1 = 2x^2 + 3/( x^2 + 1)^2 dx

14. amistre64

(S) 1/(1+tan^2(x)) dx ??

15. amistre64

wanna take lazy calc, real calc, and stats over the summer :)

16. anonymous

Okay, here it is:$\arctan(x)=y$$\tan(y)=x$$dx/dy = \sec^2(x) = tan^2(x)+1$ We want dy/dx, not dx/dy, so invert it:$dy/dx = \frac{1}{\tan^2(y)+1}$ and because $\tan^2(y) = x^2$ then$dy/dx = \frac{1}{1+x^2.}$ Now we're just integrating it, instead of differentiating the opposite.

17. anonymous

why do(S) 1/(1+tan^2(x)) dx would be???

18. amistre64

yeah :)

19. anonymous

if u work like that...thats brilliant idea quantum,,,but we still need to diferentiate 1 and 1+x^2 to solve the problem...

20. amistre64

suppose I shoulda changed the variable :)

21. anonymous

and i think the answer is 2x/(1+x^2)^2 based on Ur way...

22. anonymous

Sorry, in the third line I should have said sec^2(y) = tan^2(y)+1, sorry if that confused anyone.

23. amistre64

no more confused than usual :)

24. anonymous

We established with that, that the derivative of arctan(x) = 1/(1+x^2). When we integrate the derivative, we're going to get back to arctan(x).

25. anonymous

then tell me the idea that u're supposing tan(y)=x?

26. amistre64

the books call it "trig" substitution... if i recall correctly

27. anonymous

It is trig substitution. :P

28. anonymous

the book of Calculus?...

29. amistre64

yep; all the ones from Newton on down have had it :)

30. anonymous

@tianpradana: If tan^-1(x) = y, then x = tan(y).

31. amistre64

Leibnez even considered it..maybe :)

32. anonymous

oh cheh thanks :-)

33. amistre64

tends to be around, integration by parts, u substitutions..trig subs.... inthat general area :)

34. anonymous

brilliant idea guys...

35. amistre64

its the area I have been avoinding like the plague....