Hi guys and gals, I had this question on my cal test and for some freakin reason I couldn't figure it out. If someone knows what to do, it would help for next time I guess.
integral dy/(y^2 - 4)
I'm guessing you use the integrate by parts thing..but I could be wrong.
Anywho thanks very much :)
2007-09-29 18:37:11 · 8 answers · asked by alexk 2 in Science & Mathematics ➔ Mathematics
I'm doing it by partial fractions.
Integral dy / (y+2)(y-2)
A / (y+2) + B / (y-2) = 1 / (y+2)(y-2)
A(y-2) + B(y+2) = 1
Ay - 2A + By + 2B = 1
(A+B)y = 0
2B - 2A = 1
A + B = 0
4B = 1
B = 1/4
A = -1/4
Integral 1 / [4(y-2)] - 1 / [4(y+2)] dy
Integral 1 / 4(y-2) dy
y - 2 = u
du = dy
1/4(Integral u^-1 du) = 1/4(lnu) = 1/4(ln(y-2))
Integral 1/4(y+2) dy = 1/4(ln(y+2))
1/4(ln [(y-2)/(y+2)])
Add the constant
1/4(ln [(y-2)/(y+2))]) + C
2007-09-29 18:45:33 · answer #1 · answered by UnknownD 6 · 0⤊ 0⤋
Integral dy / (y+2)(y-2)
A / (y+2) + B / (y-2) = 1 / (y+2)(y-2)
A(y-2) + B(y+2) = 1
Ay - 2A + By + 2B = 1
(A+B)y = 0
2B - 2A = 1
A + B = 0
4B = 1
B = 1/4
A = -1/4
Integral 1 / [4(y-2)] - 1 / [4(y+2)] dy
Integral 1 / 4(y-2) dy
y - 2 = u
du = dy
1/4(Integral u^-1 du) = 1/4(lnu) = 1/4(ln(y-2))
Integral 1/4(y+2) dy = 1/4(ln(y+2))
1/4(ln [(y-2)/(y+2)])
Add the constant
1/4(ln [(y-2)/(y+2))]) + C
2007-09-30 02:04:38 · answer #2 · answered by lukey7650 2 · 1⤊ 0⤋
break 1/(y^2-4) into A/(y-2) +B/(y+2) = 1/(y^2-4). Obtaining A(y+2) +B(y-2) = 1 Hence (A+B)y +2A-2B = 1or
A + B = 0
2A-2B=1
from which 4A = 1 and A = 1/4, B = -1/4.
The integral becomes Integral[(1/4) (1/(y-2) -1/(y+2))] = 1/4[ln(y-2) -ln(y+2)] = 1/4 ln[(y-2)/(y+2)] + C
2007-09-30 01:51:38 · answer #3 · answered by guyava99 2 · 1⤊ 0⤋
Integral of dy/(y^2 - 4)
= dy/(y^2 - 2^2)
= 1/2X2 log|(2+x)/(2-x)| + C Ans.
General form:
dx/(x^2 - a^2) = 1/2a [log|(a+x)/(a-x)|] + C
:)
2007-09-30 01:51:57 · answer #4 · answered by Anonymous · 1⤊ 0⤋
You can do it by parts or separate it by partial fractions. Either way I got
(1/4)(ln(x-2) - ln(x+2)) + C
Doug
2007-09-30 01:43:56 · answer #5 · answered by doug_donaghue 7 · 1⤊ 0⤋
1 / (y - 2)(y + 2) = A / (y - 1) + B / (y + 2)
1 = A(y + 2) + B(y - 1)
1 = 2A
A = 1 / 2
1 = - 3B
B = - 1 / 3
I = (1/2) â« 1 / (y - 1) dy - (1 / 3) â« 1 / (y + 2) dy
I = (1/2) log (y - 1) - (1/3) log (y + 2) + C
2007-09-30 05:12:57 · answer #6 · answered by Como 7 · 1⤊ 0⤋
integral -->S
S(dy)/(y^2 - 4)
y=2sec u
dy=2(sec u)(tan u)du
S (2sec u)(tan u)du / (4(tan u)^2)
(1/2) S (sec u)du / (tan u)
(1/2) S(csc u)du
(1/2)Ln(csc u - ctg u)+C
2007-09-30 03:08:46 · answer #7 · answered by Anonymous · 1⤊ 0⤋
Use partial fractions
2007-09-30 01:42:37 · answer #8 · answered by Demiurge42 7 · 1⤊ 0⤋