what is the intergral of 1/1 +x^2 dy

- anonymous

what is the intergral of 1/1 +x^2 dy

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- anonymous

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

- amistre64

1/1 = 1

- amistre64

but that would be to obvious to be right :)

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## More answers

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- 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 :)

- amistre64

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

- 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....

- 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

- amistre64

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

- amistre64

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

- 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.

- anonymous

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

- amistre64

yeah :)

- 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...

- amistre64

suppose I shoulda changed the variable :)

- anonymous

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

- anonymous

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

- amistre64

no more confused than usual :)

- 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).

- anonymous

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

- amistre64

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

- anonymous

It is trig substitution. :P

- anonymous

the book of Calculus?...

- amistre64

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

- anonymous

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

- amistre64

Leibnez even considered it..maybe :)

- anonymous

oh cheh thanks :-)

- amistre64

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

- anonymous

brilliant idea guys...

- amistre64

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

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