find the intergral of 1/((1+9(x^2))^2)

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find the intergral of 1/((1+9(x^2))^2)

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What have you tried?
factor out the 9, youre going to get arctan something
alright i know what i need to do i need to use trig subsitution and i wasn't sure 1/((1+9(x^2))^4/2) or should i just use tan i mean what is going to be the co-effiecinet but that seems incorrect

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oh nevermind, i didnt see the square . you can do trig substitution
ohk i can factor out the 9 that seems a good think thanks i just wasn't doing that :| i guess
whoa, where did 4/2 come from?
4/2 is the same thing as 2 so when i see all the examples in book they have roots under the section of trig subsitution
dont use 4/2, thats weird
It is weird. Very weird. Use a tan sub, as above.
if youre using an integral table look for 1 / ( x^2 + a^2)^2 dx
yup i am using a tan sub just sure what to put as a coefficient for tan
Integral tables are for babies.
yes that i what i am following cantroset but in that the a will 1/3
errr you can have 1 / ( ax^2 + b^2) ^2 dx,
i know that formula :P
whats your table say
:) so 1/3 should be my coefficient right :)
lets use trig sub just to be thorough, i use pauls online math course, here one sec i will post link
\[\text{Let } x = \frac{1}{3}\cdot \tan \alpha\]
Paul is the man!! http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx scroll down
yes why is that a 1/3 that is what i am lost about shouldn't it be 1/9
ok one sec, i hate this thing it keeps freezing. yeesh why doesnt google help with the appearance of this
the 1/3 is squared to 1/9
ohk so it is (1+ 1/3 tan (theta))^2
(1+ 1/3 tan (theta)^2)^2
oh i think i see what youre doing, youre forcing it to be a square root ok sqrt a^2 + b^2x^2 , let x = a/b tan t
\[1+9x^2 = 1+ 9\left(\frac{1}{3}\tan \alpha \right)^2 = 1+\tan^2 \alpha\]
so you want ( 1 + 9x^2)^2 = [ (1+9x^2)^4] ^1/2 ?
i think newton thats simpler,
paul seems to say you want a square root though
oh that makes so much more sense :)
you don't really need to have it i guess you can force it but you don't need it you need it for inital i guess it won't really change anything ( i am referring to the square root)
xD how on earth does that make more sense. Crazy Americans.
you can also do partial fractions i bet
i mean what you wrote.
yeah i can do partial fraction, but i am required to do trig :)
Oh, good. And you can't split it in to partial fractions (I don't think).
here integrals involving quadratics http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithQuadratics.aspx
why not?
Because it's the same term squared on the bottom and no others.
But assuming you could put it into partial fractions, it would not simplify it at all.
so those are your factors irreducible quadratic
(9x^2+1)^2 = (tan(a)^2+1) = sec(a)^4 is this correct
but probably not the easiest approach. youre tan sub is fine , one sec
No, because it would be \[\frac{ax + b}{1+9x^2} + \frac{cx + d}{\left(1+9x^2\right)^2}\] But they all equal 0 except fro d = 1, because ... etc
avnis yes
Now include the change of what you are integrating with respect to and go from there.
ohk so next 1/sec(a)^4 = cos(a)^4
Yes, but\[\frac{\mathbb{d}\alpha}{\mathbb{d}x}\] don't forget to include
not if you use complex root factoring
what
avnis just talking to newton
Oh, OK, you can;t split it up into partial fractions in a way which would make it a retardedly different problem - is that better?
newton, it would be like partial decomp'ing 1/(x^2 + 1) = A/ (x+i) + B/ ( x -i)
i prefer approaches , one size fits all. hehe, the one size hammer that is
how do i solve the rest of it
@anvis If you are now integrating with respect something other than x, you need to account for that. Note: \[x = \frac{1}{3}\tan\alpha \implies \frac{\mathbb{d}x}{\mathbb{d}\alpha} = \frac{1}{3}\sec^2 \alpha\]
i see which will make a equal to tan^-1(3x)
so i wll be left with 1/3 integral of cos^2(a)
It will, but you don't need to include that yet. \[\int\limits \frac{1}{(1+9x^2)^2} \mathbb{d}x\] becomes \[\int\limits \frac{1}{(1+\tan^2\alpha)^2} \cdot \frac{1}{3} \sec^2 \alpha \cdot \mathbb{d}a = \int \frac{1}{3} \cos^2\alpha\ \cdot \mathbb{d}a\] yes.
Now do whatever you feel like with cos^2 to make it easier to integrate, sub back in x with the tan^(-1) thing you had above and you're done.
1/3inetgarl (1/2 (cos 2(a)+1/2) du
great job
you have an extra 1/2 there
oh yes
Sorry gang, got to go. But here's a nice one for you to try if you feel up to it: By finding constants a, b, c and d such that: \[\frac{ax+b}{x^2+2x+2} + \frac{cx+d}{x^2-2x+2} \equiv \frac{1}{x^4+4} \] Show \[\int_0^1\frac{1}{x^4+4} = \frac{1}{16}\ln 5 + \frac{1}{8}\tan^{-1}2\]
ill solve it
hmm so i got a/6 + (sin 2a)+12 and i know that a= tan^-1(3x) how i do i solve that
you make a triangle
what would the side be would one side be 3x
...
1 Attachment
that is what i got
but there is this online answer from wolfaram and it doesn't match it so i am so lost
What did you get as sin(2x) ?
2sin(x)cos(x)
OK, what what did you get as sin(x) and cos(x)? Hint, use the trinagle I uploaded, with the facts opp/hyp = sin(x) and adj/hpy = cos(x) NOTE these should all be as, NOT xs, sorry for the confusion.
ya that is why i was thinking i don't have x:|
sin(a) =3x/ sqrt (9x^2+1)
Yeah the typing on my diagram should be a, too. Indeed. And cos(a)? And then 2sin(2a) ?
sorry, only sin(2a), not 2 sin(2a)
cos (a)= 1/ sqrt(9x^2+1)
:D
got it :)
Woo
i have learnt alot today :)
Good to hear. Goodbye.
thanks :)

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