# I Integral involving square and log

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1. Apr 9, 2017

### Figaro

I have this integral that when solved, involves squares and natural logs, where $A\,$,$\,b\,$, and $\,x_e\,$ are constants while $x$ is a variable.

$A = \int_{x_e}^{x} \frac{x^2 - b^2}{x} dx = \int_{x_e}^{x} x \, dx -b^2 \int_{x_e}^{x} \frac{dx}{x} = \frac{x^2}{2} - \frac{x_e^2}{2} - b^2 \ln x + b^2 \ln x_e$

Now, I want to solve for $x$ but I can't think of a way to isolate x, maybe there is a way to integrate this another way and come up with an answer that can isolate x easily or maybe there is something I'm missing?

2. Apr 9, 2017

### aheight

When I encounter these I first check if Mathematica can solve it in terms of Product Log functions. It can. Then I try to solve it myself using the properties of Product log (or the Lambert W function). Try and investigate these functions and see if you can come up with a solution.

3. Apr 9, 2017

### Staff: Mentor

First off, it's not good practice to have the same variable as one of the limits of integration and also the variable of integration. Most textbooks would write your integral as $A = \int_{x_e}^{x} \frac{t^2 - b^2}{t} dt$. There is less chance of becoming confused.
No, there is no other way of integration that would do this. Your integral represents a function of x, and as it turns out, the function involves a log term and a term with $x^2$. There is no analytic way of isolating the variable. The best you can do is use numerical methods to approximate roots of the equation $A = \frac{x^2}{2} - \frac{x_e^2}{2} - b^2 \ln x + b^2 \ln x_e$.

4. Apr 9, 2017

### aheight

If $$y=x^2/2-b^2 \log(x)+c$$ then I believe we can solve for x using the Lambert W function. I've not done this one in particular by hand but Mathematica obtains:

$$\left\{\left\{x\to -i b \sqrt{W\left(-\frac{e^{\frac{2 c}{b^2}-\frac{2 y}{b^2}}}{b^2}\right)}\right\},\left\{x\to i b \sqrt{W\left(-\frac{e^{\frac{2 c}{b^2}-\frac{2 y}{b^2}}}{b^2}\right)}\right\}\right\}$$

Is this not an acceptable analytic solution?

5. Apr 9, 2017

### Staff: Mentor

Short of that, there's no analytic way. I was thinking about the Lambert W function when I replied earlier, but didn't mention it.