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

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• mowata
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.
mowata
TL;DR Summary
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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What's to solve? That's just an equation. No variables are defined. No values to input.

Is the question "does this have a known antiderivative"?

Substituting $t = (c/a)e^{(k-b)x}$ for $k \neq b$ leads to $$\int \frac{1}{ae^{bx} + ce^{kx}}\,dx = \frac{1}{a(k-b)}\left(\frac{a}{c}\right)^{b/(b-k)}\int \frac{t^{b/(b-k) - 1}}{1 + t}\,dt$$ which depending on the limits might be expressible in terms of complete or incomplete Beta functions with parameters $b/(b-k)$ and $-k/(b-k)$.

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mowata and berkeman
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.

SammyS
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.

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.

mowata

-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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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.

berkeman and jim mcnamara
pasmith said:
Is the question "does this have a known antiderivative"?

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

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