I'm stuck with the following problem:
Find the absolute extremes of the function $$ f\left(x,y\right)=\ln\left(4x^2+y^2+4\right)+\int _0^x\frac{2t}{t^4+1}dt $$ in the region $D$ enclosed by the ellipse $$ \frac{x^2}{1}+\frac{y^2}{4}=1. $$
What I've done so far
First we recognize
$\int _0^x\frac{2t}{t^4+1}dt=\arctan \left(x^2\right)$
So
$f\left(x,\:y\right)=ln\left(4x^2+y^2+4\right)+\arctan \:\left(x^2\right)$
If let be using lagrange's multipliers
$g(x,y)=\frac{x^2}{1}+\frac{y^2}{4}-1$
$∇f\left(x,y\right)=\lambda ∇g\left(x,y\right)$
$g(x,y)=c$
The systems of equations is the following
$\frac{8x}{4x^2+y^2+4}+\frac{2x}{x^4+1}=\lambda 2x$
$\frac{2y}{4x^2+y^2+4}=\frac{\lambda y}{2}$
$\frac{x^2}{1}+\frac{y^2}{4}=1$
And shamely I'm still dealing with it I already check the solutions in Wolfram Alpha but I want to derive it analytically or seeing how is done.
So I ask for any advice on that, and check if what I've done so far is correct. (Sorry for the basic of the question)