ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: sailinginwater on October 08, 2018, 11:09:51 pm

Title: How to find the antiderivative of 1/(1-x)? Please give detailed working
Post by: sailinginwater on October 08, 2018, 11:09:51 pm: :D
Title: Re: How to find the antiderivative of 1/(1-x)? Please give detailed working
Post by: sailinginwater on October 09, 2018, 07:46:17 am: Bump
Title: Re: How to find the antiderivative of 1/(1-x)? Please give detailed working
Post by: K888 on October 09, 2018, 10:24:25 am: Hey there :) just letting you know that you might have better luck if you post your questions in the VCE Methods Question Thread. You might even find that someone has answered a similar question before :)
Title: Re: How to find the antiderivative of 1/(1-x)? Please give detailed working
Post by: akka13722 on October 09, 2018, 11:12:27 am: I'd rewrite it as the antidiff of -(1/x-1), and then apply the relevant rule to give -ln(|x-1|) (the two lines mean you take the absolute value of x-1 i.e. ignore the negative sign if there is one and then find the natural logarithm of it).
Title: Re: How to find the antiderivative of 1/(1-x)? Please give detailed working
Post by: sailinginwater on October 09, 2018, 03:43:13 pm: Quote from: akka13722 on October 09, 2018, 11:12:27 am
I'd rewrite it as the antidiff of -(1/x-1), and then apply the relevant rule to give -ln(|x-1|) (the two lines mean you take the absolute value of x-1 i.e. ignore the negative sign if there is one and then find the natural logarithm of it).
Absolute values are taken out if the course, so is there a way to integrate it without the absolute sign?
Title: Re: How to find the antiderivative of 1/(1-x)? Please give detailed working
Post by: S_R_K on October 09, 2018, 05:24:47 pm: Quote from: sailinginwater on October 09, 2018, 03:43:13 pm
Absolute values are taken out if the course, so is there a way to integrate it without the absolute sign?

For x > 1, we have:

$\frac{d}{dx}(log_e(x - 1)) = \frac{1}{x - 1} = \frac{-1}{1 - x}$

If, on the other hand, we have x < 1, then:

$\frac{d}{dx}(log_e(-(x - 1)) = \frac{-1}{-(x - 1)} = \frac{1}{x - 1} = \frac{-1}{1 - x}$

Hence, for x > 1:

$\int \frac{1}{1 - x}dx = -\int \frac{-1}{1 - x}dx = log_e(x - 1) + c$

For the case where x < 1, we have:

$\int \frac{1}{1 - x}dx = -\int \frac{-1}{1 - x}dx = log_e(-(x - 1)) + c$

(The absolute value notation makes this simpler, because |x – 1| = x – 1 when x > 1, and |x – 1| = –(x – 1) when x < 1; so we don't have to write down two different anti-derivatives for two different parts of the maximal domain.)