integrationof log (1 Viewer)

[metalhead_loz]

I think its really unfair that we have a math assessment worth 20% on wed and our teacher has rushed through this. He did even explain integration or differentiation of log.



Show that (1+x)/(1-x)= -1 + 2/(1-x).

Hence find S (1+x)/(1-x) dx

 

[SoulSearcher]

I think I've already done this in another thread, but I'll do it again.
-1 + 2/(1-x)
= {-(1-x)+2}/(1-x)
= (-1+x+2)/(1-x)
= (1+x)/(1-x)
For the second question, sub in -1 + 2/(1-x) for (1+x)/(1-x), then integrate i.e.
S (1+x)/(1-x) dx
= S -1 + 2/(1-x) dx
= -x - 2ln(1-x) + c

 

[jake2.0]

(1+x)/(1-x) : the can be written as [-(1+x) +2]/(1-x) = -1 + 2/(1-x)

So S (1+x)/(1-x) dx = S {-1 + 2/(1-x)} dx = -x -2ln|1-x| +c

Just so u know dirivative of ln(f(x)) = f`(x)/f(x)

^he beat me 2 it

 

[metalhead_loz]

Thanks everyone

 

[metalhead_loz]

S (2x+5)/(x+4) dx

I got: 2x + 6 ln(x+4) +C

But that answer says 2x -3ln (x+4) + C


Is the answer right...I cant see where I have gone wrong. I am so mad that we had to rush through the topic. My teacher hasn't even gone through graphing them and he said he will on Tuesday, which is the day before the exam.

 

[Trev]

int. (2x+5)/(x+4) dx
int. [2 - 3/(x+4)] dx
2x - 3ln|x+4| + C ...
Did you try polynomial division?

 

[SoulSearcher]

(2x+5)/(x+4)
= (2x+8-3)/(x+4)
=2(x+4)/(x+4) - 3/(x+4)
= 2 - 3/(x+4), therefore
S (2x+5)/(x+4) dx
= S 2 - 3/(x+4) dx
= 2x - 3ln(x+4) + c

Got beaten by ^

 

[metalhead_loz]

Damn. I guess the way I have been doing it only works for some.

I split it up.

S 2x/x+4 + 5/(x+4)


SO I guess I shouldn't do that...


Edit: Do you only use that when there are no common factors?