# Problem with integration, relating natural log and arctanh

1. May 2, 2015

### alocs.3

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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. May 2, 2015

### vela

Staff Emeritus
You dropped a factor of j and have a sign error. Also, use the fact that
$$\tanh^{-1} \alpha = \frac 12 \log \frac{1+\alpha}{1-\alpha}.$$

3. May 2, 2015

### Delta²

Just devide the numerator and denominator of the fraction you got in ln(..) by $\sqrt\Delta$. Also j shouldnt 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.

Last edited: May 2, 2015
4. May 2, 2015

### alocs.3

Thanks for the help, both of you!
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:

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. May 3, 2015

### vela

Staff Emeritus
Perhaps you need to use the fact that $\frac{1+a}{1-a} = \left(\frac{1-a}{1+a}\right)^{-1}$.

6. May 3, 2015

### Delta²

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...

Last edited: May 3, 2015