Recent content by kuskus94

By differentiating it? $$ \frac {d(-ln(u + 2) + ln(u + 1) + C)}{du}$$ $$ = \frac{-1}{(u+2)} + \frac {1}{(u+1)}$$ $$ = \frac {-(u+1)+(u+2)}{(u+2)(u+1)} $$ $$ = \frac {1}{(u+2)(u+1)} $$

Okay, I am having trouble on determining the A and B. \frac {1}{(u+2)(u+1)} \frac {A}{(u+2)} + \frac {B}{(u+1)} \frac {A(u+1)+B(u+2)}{(u+2)(u+1)} A(u+1)+B(u+2)=1 There are two constants that I can not determine its value. :confused: Or is it something like this : 1 = Au+A+Bu+2B 1 =...

So, the final answer is m=|x+14|? I am still in doubt. :confused:

Oh, it was just an example... :) How can I put it into the form of log? So, the first thing I have to do is factorizing the denominator \int \frac {1}{u^2+3u+2} du=\int \frac {1}{(u+2)(u+1)} du and after that, I should get something like =\int (u+2)^{-1}(u+1)^{-1} du then, assume that (u+2)...

Hmmm... so, I need to add m>0 to the equation or something like that. m=|x+14|; m>0?

Wait, shouldn't it have to be in a form like \frac{1}{x(x-2)^2}=\frac{A}{x}+\frac{B}{(x-2)^2}+\frac{C}{x-2} O_o I still don't get it. Is the "d" coefficient or the symbol of differentiation?

First, I apologise that I have posted this question in the wrong section. I should have posted it in the precalculus section. Homework Statement \int \frac {1}{u^2+3u+2} du Homework Equations \int \frac {1}{x} dx = ln{x} We know that \frac {d(ln(u^2+3u+2))}{dx} =\frac {1}{u^2+3u+2} *...

because |m|=\frac{|(x+14)(x-10)|}{|(x-10)|}, is it |m|=|(x+14)| ?

Okay, sorry it should be |x-10|{m}=|(x+14)(x-10)| So, if I do this : (x-10)^2{m}^2=((x+14)(x-10))^2 is it possible?

Homework Statement Determine m : |x-10|<{1}/{m} if its final form is : |x^{2}+{4}x-140|<1 Homework Equations To remove the modulus, square them... The Attempt at a Solution I have tried to assume that if |x-10|{m}<{1} then, I can find |x-10|{m}=|x^{2}+{4}x-140|...

Ah, I see... Thank you very much for your help. :)

Homework Statement http://sphotos-h.ak.fbcdn.net/hphotos-ak-snc6/282312_368248176595083_1229435302_n.jpg Homework Equations To subtitute x into U, but that did not work. The Attempt at a Solution I have tried subtituting x into U like this ...

Dear lurflurf, I sill don't get how the question a) can me modified to be like that and I would like to ask you to break it down more for me. As for question b), I apologize that my lack of reading. It is my bad to not read the calculus book deeper. :blushing: Thank you for your reply...

Hi, I am a newbie here and a newbie too at calculus since I just started my first year in college. Anyway, there are some questions that I don't quite understand and I hope that anyone can help me with this. http://sphotos-a.ak.fbcdn.net/hphotos-ak-ash4/269471_368130629940171_945927165_n.jpg...