anonymous
  • anonymous
hi im having trouble finding the integral 1/u^2-16
Mathematics
katieb
  • katieb
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bahrom7893
  • bahrom7893
is u^2 -16 in the denominator? or is only u^2 in the denominator?
anonymous
  • anonymous
u^2-16 is in the denominator
bahrom7893
  • bahrom7893
okay

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bahrom7893
  • bahrom7893
workin on it
anonymous
  • anonymous
k thank you
anonymous
  • anonymous
substitue u for tan(x)
bahrom7893
  • bahrom7893
use trig substitution..
anonymous
  • anonymous
i cant because there is a minus sign
bahrom7893
  • bahrom7893
Waiy
bahrom7893
  • bahrom7893
U don't have to
bahrom7893
  • bahrom7893
integration by parts
anonymous
  • anonymous
not tangent sec
bahrom7893
  • bahrom7893
Sorry meant partial fractions
anonymous
  • anonymous
use sec(x)=u
bahrom7893
  • bahrom7893
I would go this way: 1/(u^2-16) = 1/[(u-4)(u+4)]
bahrom7893
  • bahrom7893
A/(u-4) + B/(u+4) = 1/(u^2-16) its easier.
bahrom7893
  • bahrom7893
Au + 4A + Bu - 4B = 1
anonymous
  • anonymous
i did that and i got -1/8lnu{u+4)+1/8ln(u-4) but its wrong
bahrom7893
  • bahrom7893
Au + Bu +4A - 4B = 1 (A+B)u + (A-B)4 = 0*u + 1
bahrom7893
  • bahrom7893
A+B = 0 (A-B) * 4 = 1
bahrom7893
  • bahrom7893
so now let me check over my arithmetic.. I get confused when I have to type out math problems.
anonymous
  • anonymous
k
bahrom7893
  • bahrom7893
http://www.twiddla.com/499646 hey can u go there, its easier for me to write there..
anonymous
  • anonymous
i get (1/8)ln(u-4) - (1/8)ln(u+4)

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