Re: Simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg119252] Re: Simple integral
- From: Mariano Pierantozzi <mariano.pierantozzi at gmail.com>
- Date: Thu, 26 May 2011 07:17:41 -0400 (EDT)
I am grateful to all who answered me.
All of you have suggest to me to use the command
Simplify[% // TrigToExp, b^2 - 4 c > 0]
In this way i've this solution:
(Log[-b + Sqrt[b^2 - 4*c] - 2*x] - Log[b + Sqrt[b^2 - 4*c] + 2*x])/
Sqrt[b^2 - 4*c].
But this is not the solution of the departure integral.
If "whit my hand" calculate the solution i've this result:
1(1/Sqrt[b^2 -
4 c]) (Log[(2 x +
b - (Sqrt[b^2 - 4 c]))/((2 x + b + (Sqrt[b^2 - 4 c])))])
And i thing it's not the same, because i've to do the limit x-->Infininity
and in this case the mathematica solution is, once again, in complex field.
Can you help me again?
Thank you again!
Mariano Pierantozzi
PhD Student
Energy Engineering
2011/5/23 Mariano Pierantozzi <mariano.pierantozzi at gmail.com>
> 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
>