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What have you tried?

factor out the 9, youre going to get arctan something

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?

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

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

\[1+9x^2 = 1+ 9\left(\frac{1}{3}\tan \alpha \right)^2 = 1+\tan^2 \alpha\]

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 :)

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).

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

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

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

i see which will make a equal to tan^-1(3x)

so i wll be left with 1/3 integral of cos^2(a)

1/3inetgarl (1/2 (cos 2(a)+1/2) du

great job

you have an extra 1/2 there

oh yes

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

...

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)

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 :)