2006-10-31 21:27:33 · 6 answers · asked by kasif k 1 in Science & Mathematics ➔ Mathematics
Do you mean:
∫dx/(2x+1) ?
If so then:
u=2x+1, du=2 dx
1/2 ∫1/u du
1/2 ln |u| + C
1/2 ln |2x+1| + C
2006-10-31 21:37:19 · answer #1 · answered by Pascal 7 · 0⤊ 0⤋
S means integration! right?
Sdx/(2x+1)
put, 2x+1 = p
hence, 2dx = dp
ie, dx = (1/2)dp
hence,
Sdx/(2x+1)
=S(1/2)dp/p
=(1/2) S(dp/p)
= (1/2) ln | p | + c
= (1/2) ln |2x+1| + c (ans)
2006-11-01 05:41:12 · answer #2 · answered by s0u1 reaver 5 · 0⤊ 0⤋
1\2( half of )log (2x+1) +c
2006-11-01 05:36:19 · answer #3 · answered by Adi 1 · 0⤊ 0⤋
int dx/(2x+1)
=2 {int dx/(x+1/2)}
=2 {(x+1/2)^2}/2
= {2x +1}^2/4
2006-11-03 06:45:36 · answer #4 · answered by mou 2 · 0⤊ 0⤋
its just a fraction, what about it. You cant solve it, since it isnt an eqation. it is in it's simplest form since the x is on the top can't cancel the x on the bottom since it is within the parenthesis .
2006-11-01 05:32:06 · answer #5 · answered by chance m 2 · 0⤊ 1⤋
ln(sqrt(2x+1)) + c
2006-11-04 09:48:05 · answer #6 · answered by Atul I 2 · 0⤊ 0⤋