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Re: Simple integral

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On 5/26/11 at 7:17 AM, mariano.pierantozzi@gmail.com (Mariano
Pierantozzi) 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])))])

First, consider:

In[31]:= g =
  Simplify[Integrate[1/(x^2 + b x + c), x] // TrigToExp, b^2 - 4
c > 0]

Out[31]= (Log[Sqrt[b^2 - 4*c] - b - 2*x] -
    Log[Sqrt[b^2 - 4*c] + b + 2*x])/
    Sqrt[b^2 - 4*c]

In[32]:= D[g, x] // Simplify

Out[32]= 1/(x*(b + x) + c)

So, the result obtained from Integrate is clearly the
anti-derivative of the integrand and is a correct solution to
the indefinite integral.

Now consider your hand solution

In[33]:= f =
   1 (1/Sqrt[
       b^2 - 4 c]) (Log[(2 x +
         b - (Sqrt[b^2 - 4 c]))/((2 x + b + (Sqrt[b^2 - 4 c])))]);
D[f, x] // Simplify

Out[34]= 1/(x (b+x)+c)

which demonstrates both f and g are valid solutions to the
indefinite integral. But since,

In[35]:= FindInstance[g != f, {b, c, x}]

Out[35]= {{b -> 149/10 - (11*I)/2, c -> 59/10 - (62*I)/5,
      x -> 49/2 - I/10}}

it is clear f and g are identical for all b, c and x.

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Re: Simple integral Bill Rowe <readnews@sbcglobal.net> - 2011-05-27 10:14 +0000

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