Someone want to complete the square for me, on the bottom? Never truly learned how to do it... know you have to take half the mid. term....
Integral of ((1/(9x^2 - 36x + 40)

Question

Someone want to complete the square for me, on the bottom? Never truly learned how to do it... know you have to take half the mid. term....
Integral of ((1/(9x^2 - 36x + 40)

hartnn · Answer

what is middle term ?

anonymous · Answer

I lied.... now I think it's supposed to change to 9(x -2)^2 + 4..... which doesn't help me anyway.... because I'm not sure how to get that into the 1/(x^2 +1)....

hartnn · Answer

take 9 common from 9(x -2)^2 + 4
what u get ?

anonymous · Answer

Well.... techincally I'm supposed to get the four out of there first. so it's 1/4 times the integral.... and based on notes it's almost like I would have to go 9/4(x-2)^2 + 1...... which is just.... ugh.

anonymous · Answer

I don't think I took good notes on this junk.

hartnn · Answer

ok, u can do that also.
so u have 
9/4(x-2)^2 +1
= $\huge ((2/3)(x-2))^2+1 $
now put y= 2/3(x-2)
and tell me what u get

hartnn · Answer

from 9/4
(2/3)^2 = 9/4
 i wanted to bring 9/4 inside square term

hartnn · Answer

confused ?

anonymous · Answer

just a sec

anonymous · Answer

I'm still confused on how the (2/3)^2 is equivalent to 9/4..... because.... it's not..... or else I'm just worse at math than I thought. I would have reversed the two numbers...

hartnn · Answer

omg! i id a typo, sorry, 
its (3/2)^2
sorry.

hartnn · Answer

one more typo *did

anonymous · Answer

Okay, I think I can get the right answer from there, thanks!
.... but I'll probably be right back on the site with more integral crap.

hartnn · Answer

sure do come, happy to help ^_^