2007-06-22 01:31:20 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
4 / x(x+1) = 4/x - 4/(x+1)
Int. [ 4/x - 4/(x+1) ] = 4 lnx - 4 ln(x+1)
2007-06-22 01:39:02 · answer #1 · answered by Sanny 2 · 0⤊ 0⤋
Let
4 / (x.(x + 1)) = A / x + B / (x + 1)
4 = A(x + 1) + Bx
4 = (A + B).x + A
A = 4
A + B = 0
B = - 4
I = 4 ∫ 1/x dx - 4 ∫1 / (x + 1).dx
I = 4.log x - 4.log (x + 1) + C
I = 4[ log x - log (x + 1)] + C
I = 4 log [ x / (x + 1)] + C
or
I = log [ (x / (x + 1))^4] + log K
I = log [ K.(x / (x + 1))^4]
2007-06-24 20:08:44 · answer #2 · answered by Como 7 · 0⤊ 0⤋
Absolutely, you can ....
Use cover rule :
4/x(x+1) = 4/x - 4/(x+1)
int ( 4/x(x+1) ) = int 4/x - int 4/(x+1)
= 4ln IxI - 4 ln Ix+1I + C
= 4 In I x/(x+1)I + C
2007-06-22 01:40:15 · answer #3 · answered by Maxis 2 · 0⤊ 0⤋
4/(x*(x+1)) = A/X + B/(X+1)
=> AX + A + BX = 4 => A = 4, B = -4
int 4/X -4/(X+1) = 4lnX -4LN(X+1) + C
= 4Ln((X/(X+1)) + C
2007-06-22 01:42:30 · answer #4 · answered by telsaar 4 · 0⤊ 0⤋
4/x * (x + 1)
(4x + 4) / x
4 + 4/x
2007-06-25 18:34:38 · answer #5 · answered by Jun Agruda 7 · 2⤊ 0⤋
Yes, this is an equal opportunity subject. Anyone can do it. Even you.
2007-06-22 01:37:55 · answer #6 · answered by Anonymous · 0⤊ 0⤋