abz_tech
integration question
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abz_tech
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\[\int\limits\limits_{0}^{1} \frac{ x+1 }{ x ^{2}+6x+5 }dx\]
Mimi_x3
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\[x^2+ 6x + 5 = (x+3)^2 -(3)^2 + 5 = (x+3)^2 -9 +5 = (x+3)^2 -1 \]
\[\int\limits \frac{x+1}{(x+3)^{2}-1} \]
Mimi_x3
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Wait. \[\frac{x+1}{(x+5)(x+1)} = \frac{A}{(x+5)} +\frac{B}{(x+1)} \]
Better now. :D
abz_tech
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0_o
Mimi_x3
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For A: \(x= -5\) \[\frac{(-5+1)}{(-5+1)} =\frac{-4}{-4}=1 \]
Hero
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Aren't you forgetting something @Mimi_x3 ?
Mimi_x3
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For B: \(x= -1\) => \(0\)
What am I forgetting?
Hero
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\[\frac{(x+1)}{(x+5)(x+1)} = \frac{1}{x+5}\]
Mimi_x3
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There is something wrong.
abz_tech
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isn't u's and du/dx supposed to come into play about now 0_o?
Mimi_x3
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Oh crap! Sorry!
abz_tech
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0_o
Mimi_x3
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So, now \(u= x+5\)
Hero
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\[\int\limits_{0}^{1}(x+5)^{-1}dx\]
Mimi_x3
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I was right anyway :P
just went the long way.
Hero
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Ya, exactly. You're so silly :P
Mimi_x3
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That doesn't work hero,
u = x+5
Hero
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I can make u = x + 5 if I want or not. It doesn't matter. It will still work.
abz_tech
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how would u solve (x+5)^-1
abz_tech
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\[\ln x+5?\]
winterfez
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no i guess that does not work
Mimi_x3
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\[\int\limits\frac{1}{x+5}dx \]
\[u = x+5 => \frac{du}{dx} = 1 => du = dx \]
\[\int\limits\frac{1}{u} du = \ln(x+5) + C\]
@hero: writing in that form won't work.
Hero
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\[\ln(|x+5|)\]
Hero
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I don't need @mimi_x3 ( a kid) to tell me things, lol
winterfez
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@hero right, is absolute value for ln
Mimi_x3
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I am not a kid!!
It's the truth it won't work in the form; of if you want to use that retarded formula.
abz_tech
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hero how would u go about doing \[\int\limits_{0}^{1} (x+5)^{-1}\]
Mimi_x3
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Using that form; wont get you anyway; and get lost.
abz_tech
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fine mimi, can u tell me your way =/?
Hero
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She already did. @winterfez, also provided some good steps.
Mimi_x3
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i typed up the solution!!! scroll upp!!
Hero
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I think I deserve a medal for pointing @mimi_x3 in the right direction.
abz_tech
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then with the limits would i be ln 6 - ln 5?
Mimi_x3
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Well, first I was not wrong,I just went the long way. I would have gotten the right anwer without your input.
Secondly, can you please show us how you would integrate \[\int\limits(x+5)^{-1} dx\]
Hero
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That's top secret info.
Mimi_x3
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Lol, because it's impossible :)
Mimi_x3
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Me want a medal as well!
Hero
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Nothing is impossible
Mimi_x3
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:D