dy/dx= 1 / [x^2 - 1]
y=?
2007-09-27 04:31:39 · 5 answers · asked by Chocolate Strawberries. 4 in Science & Mathematics ➔ Mathematics
dy/dx= 1 / [x^2 - 1] = (1/2) [ 1 / (x - 1) - 1 / (x + 1) ]
dy = (1/2) [ 1 / (x - 1) - 1 / (x + 1) ] dx
Integrating,
y = (1/2)[ l ln l (x - 1) l - ln l ( x + 1 ) l ]
= (1/2) ln l (x -1 ) / ( x + 1) l
2007-09-27 04:40:39 · answer #1 · answered by Madhukar 7 · 0⤊ 0⤋
=(1/2)ln[(x-1)/(x+1)] for x^2 > 1
=-(1/2)ln[(1+x)/(1+x)] for x^2 < 1
2007-09-27 11:38:45 · answer #2 · answered by supastremph 6 · 0⤊ 0⤋
1/2 * [ ln ( x-1 ) - ln ( x +1 ) ]
2007-09-27 11:38:14 · answer #3 · answered by jaguet 1 · 0⤊ 0⤋
y=integ of 1 / (x^2-1)
x=sin(t)
dx=cos(t)dt
integ cos(t)dt / (sin^2(t)-1)
integ cos(t)dt / cos^2(t)
integ sec(t)dt = ln |sec(t)+tan(t)|+C
where t=ARCSIN(x)
2007-09-27 12:07:24 · answer #4 · answered by cidyah 7 · 0⤊ 0⤋
i just answered the question somebody asked 5 min ago.
2007-09-27 11:35:37 · answer #5 · answered by Anonymous · 0⤊ 0⤋