2006-12-08 02:58:30 · 6 answers · asked by dharma&nirvana 1 in Science & Mathematics ➔ Mathematics
Write x of numerator as x-1+1
Split (x-1/ x-1) + (1 /x-1 )
=1 +1/x-1
Integrate to get,
x +log(x-1)
2006-12-08 03:02:23 · answer #1 · answered by amudwar 3 · 0⤊ 0⤋
Substitute u=x-1, you have du=dx and therefore
x/(x-1) dx = (u+1)/u du = (1 + 1/u) du
which integrates to:
u + lnu + C
= x - 1 + ln(x-1) + C
= x + ln(x-1) + C because the -1 can be absorbed into the constant term.
2006-12-08 11:05:12 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x/x-1 = (x-1 + 1)/x-1 = 1 + 1/x-1
integ(1+1/x-1)dx = x + ln|x-1| + c
2006-12-08 11:24:48 · answer #3 · answered by SaturnReLnArimani 2 · 0⤊ 0⤋
â«x/(x-1) dx =
u = (x-1) => x = (u+1)
du = dx
=>
â«x/(x-1) dx =
â«(u+1)/u du =
â«u/u du +â«1/u du =
â«du +â«1/u du = u + ln u + k
=>
â«x/(x-1) dx = (x-1) + ln(x-1) + k
â«x/(x-1) dx = x + ln(x-1) + (k-1)
â«x/(x-1) dx = x + ln(x-1) + c
₢
2006-12-08 11:02:44 · answer #4 · answered by Luiz S 7 · 0⤊ 0⤋
add andsubtract1
=x-1+1/x-1
=(1+1/x-1))dx
=dx+dx/(x-1)
integrating
=x+ln(x-1)+C
2006-12-08 11:02:56 · answer #5 · answered by raj 7 · 0⤊ 0⤋
ln(abs(x-1))+x
2006-12-08 11:02:06 · answer #6 · answered by PaD 2 · 0⤊ 0⤋