- #1

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[itex]\frac {1} {\sqrt {(k+p-c)^2 + 4 k c}-k-p+c}[/itex]

k, p and c are all real and positive.

I submitted it to Maxima, but it stayed implicit.

Can you please suggest any substitution or other technique to solve it?

Thanks

L

- Thread starter lavoisier
- Start date

- #1

- 177

- 24

[itex]\frac {1} {\sqrt {(k+p-c)^2 + 4 k c}-k-p+c}[/itex]

k, p and c are all real and positive.

I submitted it to Maxima, but it stayed implicit.

Can you please suggest any substitution or other technique to solve it?

Thanks

L

- #2

- 22,089

- 3,290

- #3

- 177

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I multiplied both numerator and denominator by the factor you said, expanded, and got this:

[itex]\frac {\sqrt {(k+p-c)^2 + 4 k c}+k+p-c} {4 k c}[/itex]

which I submitted to Maxima's integrate function, and I got the result.

Apologies for not writing it down, it's a long sum of logarithms, a square root, a linear term in c and asinh functions.

I don't exactly see how integrating this must involve inverse hyperbolic functions, except perhaps for the known relationship:

[itex]asinh(x)=Ln(\sqrt{1+x^2}+x)[/itex]

Anyway, it does the job, so...

Thanks!

L

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