Find real roots: $2\arctan x+\ln(1+x^2)-2\ln|x|=3$ How would you go about solving this equation?
2arctanx+ln(1+x^2)-2ln|x|=3
Can we use the derivative to solve it?
The problem also says "Illustrate your answer with an appropriate figure".
Thanks 
 A: You can rewrite the equation as
$$2\arctan x=3-\ln\left(1+{1\over x^2}\right)$$
The curve 
$$y=2\arctan x$$
should be easy to sketch, at least roughly:  It increases steadily from an asymptotic value of $-\pi$ for negative values of $x$ to an asymptotic value of $\pi$ for positive values of $x$.  Similarly, the curve
$$y=3-\ln\left(1+{1\over x^2}\right)$$
is not too hard to sketch roughly:  For positive values of $x$, it increases from $-\infty$ at $x=0$ to an asymptotic value of $3$ as $x\to\infty$.  For negative values of $x$, it's the mirror image across the $y$ axis.
The sketches instantly show that the two curves cross exactly once to the left of the $y$ axis.  That is, the equation has exactly one solution with $x\lt0$.  
For $x\gt0$, the curve with the logarithm obviously starts below the arctangent curve.  It also finishes below it, since the asymptotic value $3$ is less than the asymptotic value $\pi$.  To see if the two curves ever cross (i.e., if the equation has any solutions with $x\gt0$) requires a little more care.  Plugging in $x=1$, we find $2\arctan(1)=\pi/2$ is less than $3-\ln(2)$, so the two curves definitely cross at least once for $0\lt x\lt1$, and at least once again for some value $x\gt1$ (since, as we said, the arctangent curve finishes above the logarithm curve).  
In fact there are just two crossings, which you can prove by taking the derivative of the difference,
$$f(x)=2\arctan x+\ln\left(1+{1\over x^2}\right)-3\implies f'(x)={2(x-1)\over x(1+x^2)}$$
which shows the difference is decreasing for $0\lt x\lt1$ and increasing for $x\gt1$.
As for the solutions themselves, I think there's no easy way to find them.  I had a graphing program draw the aforementioned curves, and it automatically labeled the crossing points to three decimal places as $x\approx-0.188$ to the left of the $y$ axis and $x\approx0.319$ and $13.581$ for the positive values.  Of course I could have just had the program graph the difference function, but the point I'm trying to make is that, by setting things up carefully, you can get a pretty good idea of what things look like without resorting to a boatload of calculations.
A: $$2\arctan{(x)}+\ln(1+x^2)-2\ln|x|=3\Rightarrow 2\arctan(x)+\ln(1+x^2)+\ln(\frac{1}{x^2})=3$$
The left side can be rewritten as 
$$2\arctan(x)+\ln(1+x^2)+\ln(\frac{1}{x^2})=2\arctan(x)+\ln(\frac{1+x^2}{x^2})$$
Therefore 
$$2\arctan(x)+\ln(\frac{1+x^2}{x^2})=3\Rightarrow 2\arctan(x)=\ln(\frac{e^3x^2}{1+x^2})\Rightarrow x=\tan(\ln(\sqrt{\frac{e^3x^2}{1+x^2})})$$
I think at this stage you should look for a numerical analysis method to find the fixed point of the function
$$f(x)=\tan(\ln(\sqrt{\frac{e^3x^2}{1+x^2})})$$
