Doubt with an inequality involving integrals I have to prove this inequality about integrals. I did it but I'm not sure if my arguments were correct. Take a look:

Prove:
  $$0\leq\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$$

My try is this:
Let $f(x)=\frac{1-x^2}{x^4+1}$ with $f:[-1,1]\to\mathbb{R}$ and $g(x)=\frac{2+x^4}{x^4+1}$ with $g:[-1,1]\to\mathbb{R}$. So we have to prove: $f(x)\leq g(x)$ for all $x\in[-1,1]$.
Clearly we only need to prove that $1-x^2\leq2+x^4$ because the denominator is always a positive number in $[-1,1]$. So we get:
$$0\leq x^2(x^2+1)+1=(x^2+1)(x^2+\frac{1}{x^2+1})$$
Therefore we need:
$$x^2+1\geq0 \land x^2+\frac{1}{x^2+1}\geq0$$
$$\lor x^2+1\leq0 \land x^2+\frac{1}{x^2+1}\leq0$$ 
The first pair of inequalities are always true for any $x\in\mathbb{R}$, because they involve positive numbers. And the second pair is the empty set by the same reason. 
Therefore $f(x)\leq g(x)$ for any $x\in\mathbb{R}$ in particular for $x\in[-1,1]$ so we have: $\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$
The part of the zero is easy because we only need to prove: $0\leq1-x^2$, that is $x\leq|1|$, $-1\leq x\leq1$.
Is the first part correct or it is "forced"? Thanks!

$$\int\limits_{-1}^1\frac{2+x^4}{x^4+1}dx-\int\limits_{-1}^1\frac{1-x^2}{x^4+1}dx=\int\limits_{-1}^1\frac{x^4+x^2+1}{x^4+1}dx>0$$
$$\int\frac{1-x^2}{x^4+1}dx=-\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=-\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}d\left(x+\frac{1}{x}\right)=$$
$$=\frac{1}{2\sqrt2}\int\left(\frac{1}{x+\frac{1}{x}+\sqrt2}-\frac{1}{x+\frac{1}{x}-\sqrt2}\right)d\left(x+\frac{1}{x}\right)=$$
$$=\frac{1}{2\sqrt2}\ln\frac{x+\frac{1}{x}+\sqrt2}{x+\frac{1}{x}-\sqrt2}\frac{1}{2\sqrt2}+C=\frac{1}{2\sqrt2}\ln\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}+C.$$
Thus, $$\int\frac{1-x^2}{x^4+1}dx=\frac{1}{\sqrt2}\ln\frac{2+\sqrt2}{2-\sqrt2}=\frac{1}{\sqrt2}\ln\frac{\sqrt2+1}{\sqrt2-1}=\sqrt2\ln(\sqrt2+1)>0.$$
