# Integration of ∫1/(1-x)dx

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#1
Why is1/(1-x)dx
= -In|1-x|

0
6 years ago
#2
Why is1/(1-x)dx
= -In|1-x|

Nah because you have to use chain rule

If |1-x| > 0 then |1-x| = 1-x and then the derivative of ln|1-x| is the derivative of ln(1-x) which is -1/(1-x) right? Similarly if |1-x| < 0 then |1-x| = x - 1 and then the derivative is 1/(x-1)= -1/(1-x). So in any case you need to multiply by -1 to get 1/(1-x)
1
#3
(Original post by 13 1 20 8 42)
Nah because you have to use chain rule

If |1-x| > 0 then |1-x| = 1-x and then the derivative of ln|1-x| is the derivative of ln(1-x) which is -1/(1-x) right? Similarly if |1-x| < 0 then |1-x| = x - 1 and then the derivative is 1/(x-1)= -1/(1-x). So in any case you need to multiply by -1 to get 1/(1-x)
I still don't get it .. But when it comes to integrating don't we use the reverse chain rule not the chain rule for differentiation.
0
6 years ago
#4
∫ f'(x)/f(x) dx = ln(x) + c

therefore to make it represent the form ∫ f'(x)/f(x) dx
do: -∫ f'(x)/f(x) dx = -∫ -1/(1-x)dx = -ln(1-x) + c

you could use the variable u (substitution) if you like:
u=(1-x)
du/dx = -1
∫ (1/u)(-1) du = -∫ (1/u) du = -ln(u) + c = -ln(1-x) + c
1
#5
(Original post by XOR_)
∫ f'(x)/f(x) dx = ln(x) + c

therefore to make it represent the form ∫ f'(x)/f(x) dx
do: -∫ f'(x)/f(x) dx = -∫ -1/(1-x)dx = -ln(1-x) + c

you could use the variable u (substitution) if you like:
u=(1-x)
du/dx = -1
∫ (1/u)(-1) du = -∫ (1/u) du = -ln(u) + c = -ln(1-x) + c
I see thank you!!
0
2 years ago
#6
Intg of 1/x is log(x)And intg of 1/(ax b) is log(ax b)/a As (1-x) is a composite function so the (-) sign
0
2 years ago
#7
(Original post by Ajljune)
Intg of 1/x is log(x)And intg of 1/(ax b) is log(ax b)/a As (1-x) is a composite function so the (-) sign
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