Question

hi im having trouble finding the integral 1/u^2-16

Answer 1

is u^2 -16 in the denominator? or is only u^2 in the denominator?

Answer 2

u^2-16 is in the denominator

Answer 3

okay

Answer 4

workin on it

Answer 5

k thank you

Answer 6

substitue u for tan(x)

Answer 7

use trig substitution..

Answer 8

i cant because there is a minus sign

Answer 9

Waiy

Answer 10

U don't have to

Answer 11

integration by parts

Answer 12

not tangent sec

Answer 13

Sorry meant partial fractions

Answer 14

use sec(x)=u

Answer 15

I would go this way:
1/(u^2-16) = 1/[(u-4)(u+4)]

Answer 16

A/(u-4) + B/(u+4) = 1/(u^2-16)
its easier.

Answer 17

Au + 4A + Bu - 4B = 1

Answer 18

i did that and i got -1/8lnu{u+4)+1/8ln(u-4) but its wrong

Answer 19

Au + Bu +4A - 4B = 1
(A+B)u + (A-B)4 = 0*u + 1

Answer 20

A+B = 0
(A-B) * 4 = 1

Answer 21

so now let me check over my arithmetic.. I get confused when I have to type out math problems.

Answer 22

k

Answer 23

http://www.twiddla.com/499646
hey can u go there, its easier for me to write there..

Answer 24

i get (1/8)ln(u-4) - (1/8)ln(u+4)