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Mathematics 19 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

1/1 = 1

OpenStudy (amistre64):

but that would be to obvious to be right :)

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (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 :)

OpenStudy (amistre64):

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

OpenStudy (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....

OpenStudy (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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (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.

OpenStudy (anonymous):

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

OpenStudy (amistre64):

yeah :)

OpenStudy (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...

OpenStudy (amistre64):

suppose I shoulda changed the variable :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

no more confused than usual :)

OpenStudy (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).

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

It is trig substitution. :P

OpenStudy (anonymous):

the book of Calculus?...

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

Leibnez even considered it..maybe :)

OpenStudy (anonymous):

oh cheh thanks :-)

OpenStudy (amistre64):

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

OpenStudy (anonymous):

brilliant idea guys...

OpenStudy (amistre64):

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

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