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

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

0_o

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

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!

0_o

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

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

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

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

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!!

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

That's top secret info.

Lol, because it's impossible :)

Me want a medal as well!

Nothing is impossible

:D