# P3 integration (small short question)Watch

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#1
∫ ln(x-1) dx

thanks
0
14 years ago
#2
(Original post by swishmasta)
∫ ln(x-1) dx

thanks

∫ ln (x-1) dx

use u=x-1
du/dx = 1
du = dx

∫ ln(x-1)dx = ∫ ln u du
= ulnu - u + C
= (x-1)ln(x-1) - (x-1) + C
= (x-1)(ln(x-1) -1 ) + C
0
14 years ago
#3
(Original post by swishmasta)
∫ ln(x-1) dx

thanks
oh sorry i didnt see the "doing it by parts, not substitution"
0
#4
i did mention that dont use substitution, i needed to know how to do it by parts... thanks for the quick reply though, very much appreciated
0
14 years ago
#5
(Original post by swishmasta)
∫ ln(x-1) dx

thanks

∫ ln(x-1) dx = ∫ 1.ln(x-1) dx

= x.ln(x-1) - ∫x/(x-1) dx

= x.ln(x-1) - ∫(1+1/(x-1)) dx

= x.ln(x-1) - x - ln(x-1) + C
0
14 years ago
#6
Actually, he used a sub and then did it by parts, which is basically the same.

∫ ln(x-1) dx
u=ln(x-1), dv/dx=1
du/dx=1/(x-1), v=x
∫ ln(x-1) dx = xln(x-1) - ∫ x/(x-1) dx = xln(x-1) - ∫ [1 + 1/(x-1)] dx = xln(x-1) - x - ln(x-1) + C

If you really want to avoid doing any substitution.
0
#7
thanks guys , made my day
0
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