Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > comp.soft-sys.math.mathematica > #2750

Re: Simple integral

Show all headers | View raw

expr = 1/(x^2 + b x + c);

i1 = Integrate[expr, x];

i2 = Simplify[i1 // TrigToExp, b^2 - 4 c > 0]

(Log[-b + Sqrt[b^2 - 4*c] - 2*x] - Log[b + Sqrt[b^2 - 4*c] + 2*x])/
 Sqrt[b^2 - 4*c]

The result that you obtained by hand:

i3 = 1 (1/Sqrt[
      b^2 - 4 c]) (Log[(2 x + 
        b - (Sqrt[b^2 - 4 c]))/((2 x + b + (Sqrt[b^2 - 4 c])))]);

i2 and i3 differ by only a constant (the constant of integration).

D[i2, x] == D[i3, x] == expr // Simplify

True

Limit[i2, x -> Infinity]

(I*Pi)/Sqrt[b^2 - 4*c]

Subtracting this constant (relative to x) from i2 is equivalent to dividing the argument of its first Log by -1 which is the result that you obtained by hand.

% == Log[-1]/Sqrt[b^2 - 4*c]

True


Bob Hanlon

---- Mariano Pierantozzi <mariano.pierantozzi@gmail.com> wrote: 

=============
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@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
>

Back to comp.soft-sys.math.mathematica | Previous | Next | Find similar | Unroll thread

Thread

Re: Simple integral Bob Hanlon <hanlonr@cox.net> - 2011-05-27 10:12 +0000

csiph-web