Re: Simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg119192] Re: Simple integral
- From: Stefan Salanski <wutchamacallit27 at gmail.com>
- Date: Tue, 24 May 2011 05:59:21 -0400 (EDT)
- References: <irdctv$3tq$1@smc.vnet.net>
On May 23, 6:29 am, Mariano Pierantozzi
<mariano.pieranto... at gmail.com> wrote:
> Hi,
> I've got some problem studing this simple integral:
> Integrate[1/(x^2 + b x + c), x].
> The Mathematica solution is:
> (2 ArcTan[(b + 2 x)/Sqrt[-b^2 + 4 c]])/Sqrt[-b^2 + 4 c]
>
> The problem is that my secon order polinomial have two real solutions, so my
> delta (-b^2 + 4 c) is greater than zero. In this case the denominator of the
> solution does not exist or exist in complex field, but my x is a volume.. .
> I try in this way
> Integrate[1/(x^2 + b x + c), x, Assumptions -> {-b^2 + 4 c < 0}], but
> nothing!
> I can't trasform my arcotangent in two log.
> In summary I would like to have such a solution:
> Integrate[1/(x^2 + 5 x + 6), x]
> Log[2 + x] - Log[3 + x]
> but in general form.
> Sorry for my english!
> Mariano Pierantozzi
> PhD Student
> Energy Engineering
Mariano,
It turns out that since the imaginary quantity Sqrt[-b^2+4c] appears
in both the denominator, AND in ArcTan in your numerator, they cancel
out and you are left with a real answer whenever you actually evaluate
it, so plugging in numbers b and c you will get real answers all the
time (unless your bounds of integration include either of the two
poles of this equation, then you get errors because it blows up).
The relation is between arctan and arctanh, the hyperbolic tangent
inverse.
ArcTan[ x * I ] = I * ArcTanh[x]
what we can do to simplify your result is to use this relation to
cancel the I's manually, i get
-(ArcTanh[(b + 2 x)/Sqrt[b^2 - 4 c]]/(2 Sqrt[b^2 - 4 c]))
then TrigToExp gives
Log[1 - (b + 2 x)/Sqrt[b^2 - 4 c]]/(4 Sqrt[b^2 - 4 c]) - Log[1 + (b +
2 x)/Sqrt[b^2 - 4 c]]/(4 Sqrt[b^2 - 4 c])
which I believe is the general case.
-Stefan S