integration question

- anonymous

integration question

- katieb

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- anonymous

\[\int\limits\limits_{0}^{1} \frac{ x+1 }{ x ^{2}+6x+5 }dx\]

- Mimi_x3

\[x^2+ 6x + 5 = (x+3)^2 -(3)^2 + 5 = (x+3)^2 -9 +5 = (x+3)^2 -1 \]
\[\int\limits \frac{x+1}{(x+3)^{2}-1} \]

- Mimi_x3

Wait. \[\frac{x+1}{(x+5)(x+1)} = \frac{A}{(x+5)} +\frac{B}{(x+1)} \]
Better now. :D

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## More answers

- anonymous

0_o

- Mimi_x3

For A: \(x= -5\) \[\frac{(-5+1)}{(-5+1)} =\frac{-4}{-4}=1 \]

- Hero

Aren't you forgetting something @Mimi_x3 ?

- Mimi_x3

For B: \(x= -1\) => \(0\)
What am I forgetting?

- Hero

\[\frac{(x+1)}{(x+5)(x+1)} = \frac{1}{x+5}\]

- Mimi_x3

There is something wrong.

- anonymous

isn't u's and du/dx supposed to come into play about now 0_o?

- Mimi_x3

Oh crap! Sorry!

- anonymous

0_o

- Mimi_x3

So, now \(u= x+5\)

- Hero

\[\int\limits_{0}^{1}(x+5)^{-1}dx\]

- Mimi_x3

I was right anyway :P
just went the long way.

- Hero

Ya, exactly. You're so silly :P

- Mimi_x3

That doesn't work hero,
u = x+5

- Hero

I can make u = x + 5 if I want or not. It doesn't matter. It will still work.

- anonymous

how would u solve (x+5)^-1

- anonymous

\[\ln x+5?\]

- anonymous

no i guess that does not work

- Mimi_x3

\[\int\limits\frac{1}{x+5}dx \]
\[u = x+5 => \frac{du}{dx} = 1 => du = dx \]
\[\int\limits\frac{1}{u} du = \ln(x+5) + C\]
@hero: writing in that form won't work.

- Hero

\[\ln(|x+5|)\]

- Hero

I don't need @mimi_x3 ( a kid) to tell me things, lol

- anonymous

@hero right, is absolute value for ln

- Mimi_x3

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.

- anonymous

hero how would u go about doing \[\int\limits_{0}^{1} (x+5)^{-1}\]

- Mimi_x3

Using that form; wont get you anyway; and get lost.

- anonymous

fine mimi, can u tell me your way =/?

- Hero

She already did. @winterfez, also provided some good steps.

- Mimi_x3

i typed up the solution!!! scroll upp!!

- Hero

I think I deserve a medal for pointing @mimi_x3 in the right direction.

- anonymous

then with the limits would i be ln 6 - ln 5?

- Mimi_x3

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

That's top secret info.

- Mimi_x3

Lol, because it's impossible :)

- Mimi_x3

Me want a medal as well!

- Hero

Nothing is impossible

- Mimi_x3

:D

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