Problem with integration, relating natural log and arctanh

In summary, the conversation revolved around solving a differential equation in terms of natural log and inverse hyperbolic tangent function. The original solution obtained involved a sign error and a missing factor of j. It was then suggested to use the fact that arctanh is equal to half the natural log of the fraction (1+alpha)/(1-alpha). However, there was some confusion about how to apply this identity, leading to further discussion and attempts at solving the problem. Ultimately, it was found that the original answer was correct, but it needed to be rewritten in terms of inverse hyperbolic tangent. There was some confusion about the notation and sign of the final answer, but it was eventually resolved by recognizing that the original solution was actually equal
  • #1
alocs.3
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Homework Statement


The problem is actually from chemical kinetics, but my problem is solving the differential equation obtained.

(dx(t))/(dt) = -j*x^2-k*x+k*a ; x is a function of t, and j,k,a are all real positive constants.

Homework Equations


I know this is a Ricatti type equation. But this is from a class for chemists who haven't taken any differential equations classes. So I was trying to solve it by separation of variables. So, the x integral you obtain is

∫ dx/ (-j*x^2-k*x+k*a) with x from 0 to x.

The Attempt at a Solution


I integrate this by means of partial fraction decomposition to obtain the answer in terms of natural log, and in terms of Δ (the discriminant of the second order polynomial)

∫=(j/√Δ)ln[(2jx+k-√Δ)/(2jx+k+√Δ)] + c (this is the indefinite integral)

but the answer expected to obtain is in terms of arctanh (inverse hyperbolic tangent function), I know they're related by arctanhx = ln(1+x/1-x)/2 , but I can't see how to relate my answer to this.

I attach basically the same information I already wrote but in paper, and the expected solution for the differential equation, even though I know that if I can relate the ln answer to the arctan one the rest of the problem is just solving for x.
Thanks for the help!
 

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  • #2
You dropped a factor of j and have a sign error. Also, use the fact that
$$\tanh^{-1} \alpha = \frac 12 \log \frac{1+\alpha}{1-\alpha}.$$
 
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  • #3
alocs.3 said:
but the answer expected to obtain is in terms of arctanh (inverse hyperbolic tangent function), I know they're related by arctanhx = ln(1+x/1-x)/2 , but I can't see how to relate my answer to this.

Just divide the numerator and denominator of the fraction you got in ln(..) by ##\sqrt\Delta##. Also j shouldn't be there , it is just ##\frac{1}{\sqrt\Delta}ln(...)##. ( I guess you forgot to put 1/j when taking the integral of ##\frac{1}{2jx+k+\sqrt\Delta}##)

In cases like this i ve to say wolfram is your best friend, just go www.wolframalpha.com and enter " integral dx/ (-j*x^2-k*x+k*a) " in the text box to see the answer. Ofcourse one should try to work out things by himself first, but its always good to know the answer if you know how to handle it properly.
 
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  • #4
Thanks for the help, both of you!
Delta² said:
Just divide the numerator and denominator of the fraction you got in ln(..) by ##\sqrt\Delta##.

I did checked Wolfram's answer, and it is in terms of arctanh, that's why I'm trying to relate my answer to it. Also, I tried what you said, dividing by ##\sqrt\Delta## but that gives something of the form ln(x-1/x+1) and the identity is in the form:

vela said:
Also, use the fact that
$$\tanh^{-1} \alpha = \frac 12 \log \frac{1+\alpha}{1-\alpha}.$$

is there an identity I'm missing for logarithms?

I can see I'm almost there, but I don't know how to change the sign inside the log term.
 
  • #5
Perhaps you need to use the fact that ##\frac{1+a}{1-a} = \left(\frac{1-a}{1+a}\right)^{-1}##.
 
  • #6
There is a sign error afterall, as vela pointed out early on, you actually factorized jx^2+kx-ka the way you did it. So, we must put a minus sign in front so it becomes after all (using -lny=ln(1/y), ##ln\frac{2jx+k+\sqrt\Delta}{2jx+k-\sqrt\Delta}## but hmm this last expression is equal to ##2coth^{-1}(\frac{2jx+k}{\sqrt\Delta})=2tanh^{-1}(\frac{\sqrt\Delta}{2jx+k})##. i wonder what is going on here...
 
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1. What is the problem with integrating natural log and arctanh functions?

The problem with integrating natural log and arctanh functions is that they do not have simple antiderivatives. This means that it is not possible to find a single function that, when differentiated, will result in the original natural log or arctanh function.

2. Can the problem with integration of natural log and arctanh be solved?

Yes, the problem can be solved by using different methods such as integration by parts or substitution. However, these methods may result in more complicated expressions, and the integration may not always be possible.

3. What is the relationship between natural log and arctanh functions?

The natural log and arctanh functions are inverses of each other. This means that if you take the natural log of a number, and then take the arctanh of that result, you will get back the original number. This relationship is also known as the hyperbolic function identity.

4. Why is it important to understand the problem with integration of natural log and arctanh?

Understanding the problem with integration of natural log and arctanh is important for solving complex mathematical problems and for applications in fields such as physics, engineering, and statistics. It also helps in understanding the concept of inverse functions and their properties.

5. Are there any real-world applications of the natural log and arctanh functions?

Yes, the natural log and arctanh functions have various real-world applications. They are commonly used in statistics to model data, in physics to describe the behavior of certain systems, and in engineering to solve differential equations. They are also used in finance, biology, and other fields.

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