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integration question

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\[\int\limits\limits_{0}^{1} \frac{ x+1 }{ x ^{2}+6x+5 }dx\]
\[x^2+ 6x + 5 = (x+3)^2 -(3)^2 + 5 = (x+3)^2 -9 +5 = (x+3)^2 -1 \] \[\int\limits \frac{x+1}{(x+3)^{2}-1} \]
Wait. \[\frac{x+1}{(x+5)(x+1)} = \frac{A}{(x+5)} +\frac{B}{(x+1)} \] Better now. :D

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Other answers:

For A: \(x= -5\) \[\frac{(-5+1)}{(-5+1)} =\frac{-4}{-4}=1 \]
Aren't you forgetting something @Mimi_x3 ?
For B: \(x= -1\) => \(0\) What am I forgetting?
\[\frac{(x+1)}{(x+5)(x+1)} = \frac{1}{x+5}\]
There is something wrong.
isn't u's and du/dx supposed to come into play about now 0_o?
Oh crap! Sorry!
So, now \(u= x+5\)
I was right anyway :P just went the long way.
Ya, exactly. You're so silly :P
That doesn't work hero, u = x+5
I can make u = x + 5 if I want or not. It doesn't matter. It will still work.
how would u solve (x+5)^-1
\[\ln x+5?\]
no i guess that does not work
\[\int\limits\frac{1}{x+5}dx \] \[u = x+5 => \frac{du}{dx} = 1 => du = dx \] \[\int\limits\frac{1}{u} du = \ln(x+5) + C\] @hero: writing in that form won't work.
I don't need @mimi_x3 ( a kid) to tell me things, lol
@hero right, is absolute value for ln
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.
hero how would u go about doing \[\int\limits_{0}^{1} (x+5)^{-1}\]
Using that form; wont get you anyway; and get lost.
fine mimi, can u tell me your way =/?
She already did. @winterfez, also provided some good steps.
i typed up the solution!!! scroll upp!!
I think I deserve a medal for pointing @mimi_x3 in the right direction.
then with the limits would i be ln 6 - ln 5?
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 \[\int\limits(x+5)^{-1} dx\]
That's top secret info.
Lol, because it's impossible :)
Me want a medal as well!
Nothing is impossible

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