The Student Room Group
Reply 1
Original post by Adorable98
Why is 1/(1-x)dx
= -In|1-x|

I thought the answer would be In|1-x| without the minus sign in front? :redface:


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)
Reply 2
Original post by 1 8 13 20 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:redface: .. But when it comes to integrating don't we use the reverse chain rule not the chain rule for differentiation.
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
(edited 7 years ago)
Reply 4
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!!
Reply 5
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
Reply 6
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

This thread is 3 years old...

Latest