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

[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 :)

$$\int{\frac{1}{a\cdot e^{bx}+c\cdot e^{kx}}dx}$$

 

[DaveC426913]

What's to solve? That's just an equation. No variables are defined. No values to input.

 

[pasmith]

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

Substituting t = (c/a)e^{(k-b)x} for k \neq b leads to <br /> \int \frac{1}{ae^{bx} + ce^{kx}}\,dx =<br /> \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).

 

[Mark44]

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.

 

[Mark44]

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.

 

[DaveC426913]

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.

 

[Bill Simpson]

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

 

[mowata]

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.

 

[robphy]

Right-clicking the OP's MathJax and copying TeX to clipboard,
then directly pasting into WolframAlpha yields
https://www.wolframalpha.com/input?i=\int{\frac{1}{a\cdot+e^{bx}+c\cdot+e^{kx}}dx}
(which is the result of @Bill Simpson )

Using the transformed expression of @pasmith ,
https://www.wolframalpha.com/input?i=+\frac{1}{a(k-b)}\left(\frac{a}{c}\right)^{b/(b-k)}\int+\frac{t^{b/(b-k)+-+1}}{1+++t}\,dt

From a search:
https://www.searchonmath.com/result...^{kx}} dx \) hypergeometric #t=1674788392750
found this possibly useful discussion
https://math.stackexchange.com/ques...1+x^n}dx$-for-$n\in\mathbb-r$#comment-4106115

 

[mowata]

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

Substituting t = (c/a)e^{(k-b)x} for k \neq b leads to <br /> \int \frac{1}{ae^{bx} + ce^{kx}}\,dx =<br /> \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.