Integration of ∫1/(1-x)dx Watch

Adorable98
Badges: 15
#1
Report Thread starter 3 years ago
#1
Why is1/(1-x)dx
= -In|1-x|

I thought the answer would be In|1-x| without the minus sign in front?
0
reply
math42
Badges: 20
Rep:
?
#2
Report 3 years ago
#2
(Original post by Adorable98)
Why is1/(1-x)dx
= -In|1-x|

I thought the answer would be In|1-x| without the minus sign in front?
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
reply
Adorable98
Badges: 15
#3
Report Thread starter 3 years ago
#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
reply
XOR_
Badges: 20
Rep:
?
#4
Report 3 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
0
reply
Adorable98
Badges: 15
#5
Report Thread starter 3 years ago
#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
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts

University open days

  • Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 31 Jul '19
  • Staffordshire University
    Postgraduate open event - Stoke-on-Trent campus Postgraduate
    Wed, 7 Aug '19
  • University of Derby
    Foundation Open Event Further education
    Wed, 7 Aug '19

Are cats selfish

Yes (105)
58.01%
No (76)
41.99%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise