Can you help to solve this integral? (resin viscosity research)

In summary, the tex was not rendering, so the OP winged it. Without any additional information about the variables, the integral is-Hypergeometric2F1[1,-k/(b-k),(b-2*k)/(b-k),-(a*E^((b-k)*x))/c]/(c*E^(k*x)*k)and I don't find any simplifications for that, additional domain info might or might not help.
  • #1
mowata
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I have tried WolfarmAlpha but it could help me. Please note this is not a homework exercise. I am a researcher and I am looking to model viscosity development of resin. there I came across with this express :)
I have tried WolfarmAlpha but it could help me. Please note this is not a homework exercise. I am a researcher and I am looking to model viscosity development of resin. there I came across with this express :)

$$\int{\frac{1}{a\cdot e^{bx}+c\cdot e^{kx}}dx}$$
 
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  • #2
What's to solve? That's just an equation. No variables are defined. No values to input.
 
  • #3
Is the question "does this have a known antiderivative"?

Substituting [itex]t = (c/a)e^{(k-b)x}[/itex] for [itex]k \neq b[/itex] leads to [tex]
\int \frac{1}{ae^{bx} + ce^{kx}}\,dx =
\frac{1}{a(k-b)}\left(\frac{a}{c}\right)^{b/(b-k)}\int \frac{t^{b/(b-k) - 1}}{1 + t}\,dt[/tex] which depending on the limits might be expressible in terms of complete or incomplete Beta functions with parameters [itex]b/(b-k)[/itex] and [itex]-k/(b-k)[/itex].
 
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  • #4
DaveC426913 said:
What's to solve? That's just an equation. No variables are defined. No values to input.
The integral that the OP wrote is NOT an equation -- an equation states the equality of two or more expressions, where the expressions are separated by '=' symbols.
 
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  • #5
mowata said:
I have tried WolfarmAlpha but it could help me. Please note this is not a homework exercise. I am a researcher and I am looking to model viscosity development of resin. there I came across with this express :) $$\int{\frac{1}{a\cdot e^{bx}+c\cdot e^{kx}}dx}$$
It would be helpful if you told us where you found this integral.
 
  • #6
Mark44 said:
The integral that the OP wrote is NOT an equation -- an equation states the equality of two or more expressions, where the expressions are separated by '=' symbols.
Yeah. In its first iteration, the tex wasn't even rendering, so I was winging it.
 
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  • #7
Without any additional information about the variables, the integral is

-Hypergeometric2F1[1,-k/(b-k),(b-2*k)/(b-k),-(a*E^((b-k)*x))/c]/(c*E^(k*x)*k)

and I don't find any simplifications for that, additional domain info might or might not help.

Integrating from 0 to t gives

(E^(k*t)*Hypergeometric2F1[1,-k/(b-k),2+b/(-b+k),-a/c]-
Hypergeometric2F1[1,-k/(b-k),2+b/(-b+k),-(a*E^((b-k)*t))/c])/(c*E^(k*t)*k)

https://reference.wolfram.com/language/ref/Hypergeometric2F1.html
 
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  • #8
DaveC426913 said:
Yeah. In its first iteration, the tex wasn't even rendering, so I was winging it.
its just an integral, I need solution of the integral mean its antiderivative if possible. Some how tex is not rendering that's why it seems strange, the expression for the integral is dx/(a.e^(bx)+c.e^(kx)), where a, b, c and k are constants, x is the variable which is basically time, the limits of the integral goes from 0 to t.
 
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  • #10
pasmith said:
Is the question "does this have a known antiderivative"?

Substituting [itex]t = (c/a)e^{(k-b)x}[/itex] for [itex]k \neq b[/itex] leads to [tex]
\int \frac{1}{ae^{bx} + ce^{kx}}\,dx =
\frac{1}{a(k-b)}\left(\frac{a}{c}\right)^{b/(b-k)}\int \frac{t^{b/(b-k) - 1}}{1 + t}\,dt[/tex] which depending on the limits might be expressible in terms of complete or incomplete Beta functions with parameters [itex]b/(b-k)[/itex] and [itex]-k/(b-k)[/itex].
This is interesting, Let me go through this first. Thanks a lot.
 

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