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# P5 - Integration watch

1. Can anyone help me integrate this:

(x+2)/(x²+6x+4)^½

the answer is [(x+3)²-5]^½ - arcosh[(x+3)/5^½] + c

I can see where the arcosh comes from with completing the square and all but I don't know how you go about getting there. Do you split it up and do it by parts? Help me!
2. Let u = x^2 + 6x + 4
du/dx = 2x + 6

1/2.du - 1 = x+2 which is the numerator, so

The integral becomes
Int (1/(2u^(1/2)) du - Int 1/((x+3)^2 - 5) dx

I think.
3. I'll try it right now! Thanks
4. another method:
write the integral as:
(x+3-1)/(x²+6x+4)^½
which is ∫(x+3)/(x²+6x+4)^½ plus ∫-1/(x²+6x+4)^½
the second one u can do via the completing square and arcosh formula
in the first one, 2x+6 is the deravative of the bracket in the bottem, so try
differentiating (x²+6x+4)^½, and u'll get the integral.

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Updated: June 6, 2004
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