How to find the antiderivative of 1/(1-x)? Please give detailed working

[sailinginwater]

 

[sailinginwater]

Bump

[K888]

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

[akka13722]

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).

[sailinginwater]

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?

[S_R_K]

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:



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



Hence, for x > 1:



For the case where x < 1, we have:



(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.)