2006-10-07 21:25:32 · 7 answers · asked by Rrrrr... 2 in Computers & Internet ➔ Other - Computers
it is (1-x/1+x)
2006-10-07 21:37:17 · update #1
1-x/1+x
= (1-x+1-1)/(1+x)
={2-(1+x)}/1+x
={2/1+x}-1
whose integral will be
2log(1+x) -x + C
(C is a constant)
2006-10-07 22:10:03 · answer #1 · answered by aldehyde 1 · 2⤊ 0⤋
1. Integral of 1/(1+x) is ln(1+x)
2. x/(1+x) = 1-1/(1+x)
3. Integral of x/(1+x) = integral of (1-1/(1+x)) = integral of 1 - integral of (1/(1+x))
4. Integral of 1 is x
5. Putting 1, 3, 4 together, integral of (1-x)/(1+x) = ln(1+x) - (x-ln(1+x)) = 2 ln(x) - x. Plus a constant of integration.
You might be interested in:
http://www.calc101.com/webMathematica/integrals.jsp
2006-10-07 22:17:42 · answer #2 · answered by Sangmo 5 · 0⤊ 0⤋
Make sure you include all brackets in the equation. Your equation as written would translate to integral of 1, which is x + any arbritray constant. However, if there are brackets (), the integral would be different. Maybe requiring a special technique called "integration by parts".
Hope this helps.
b.
2006-10-07 21:30:34 · answer #3 · answered by Radiosonde 5 · 0⤊ 0⤋
integral 1-x/1+x dx when x goes to infinite is
as integral -x/x dx = integral -1 dx = -x = -infinit
integral 1-x/1+x dx when x goes to zero is
as integral 1/1 dx = integral 1 dx = x = 0
But IN GENERAL
integral (1-x/1+x) dx= integral (1/1+x) dx - integral (x/1+x) dx = integral (1/1+x) dx - integral (1 - (1/1+x)) dx = 2 integral (1/1+x) dx- integral (1) dx = 2 Ln (x+1) - x
ln is neperian logarithm
2006-10-07 21:41:05 · answer #4 · answered by Nazanin B 2 · 0⤊ 0⤋
Text book
2006-10-07 21:27:01 · answer #5 · answered by jn l 2 · 0⤊ 0⤋
Its bit difficult but can refer mathematics book by M.L. Khanna
2006-10-07 21:28:38 · answer #6 · answered by Cool Guy 2 · 0⤊ 0⤋
ask a physics teacher
2006-10-08 00:45:12 · answer #7 · answered by debikachakraborty 3 · 0⤊ 0⤋