Integration by parts and $dx$ notation Please overview this integral evaluation:
$$ \int x^3 \arctan(x^2)dx = \frac{x^4}{4}\arctan(x^2) - \int \frac{1}{1+x^4}2x dx $$
Let's evaluate the right term:
$$\int \frac{1}{1+x^4}\color{Blue}{2x dx} = \int \frac{1}{1+x^4}\color{Blue}{dx^2} = \int y^2 \frac{1}{1+y^2} \color{Red}{dy} = ...$$
Can you explain the legitimacy of those equalities?   
Yeah, I know that $(x^2)' = 2x$, but at the middle, we have the term $\frac{1}{1+x^4}$ stands by itself attached to $dx^2$. What's the meaning of that? And at the right we have $dy$ popped out of the blue (Funny, because I colored it blue).

Hint:
$$\int\frac{2x}{1+x^4}\,dx$$
can be evaluated through partial fraction decomposition, i.e. by noticing that:
$$\frac{1}{1-\sqrt{2}\,x+x^2}-\frac{1}{1+\sqrt{2}\,x+x^2} = \frac{2\sqrt{2}\,x}{1+x^4}.$$
The antiderivatives of both terms in the LHS is an arctangent. 
However, by noticing that $x\to\sqrt{x}$ is a smooth bijective map on $\mathbb{R}^+$, a change of variable ($x=\sqrt{u}$, so that $dx=\frac{1}{2\sqrt{u}}\,du$) gives:
$$ \int_{0}^{t}\frac{2x}{1+x^4}\,dx = \int_{0}^{t^2}\frac{du}{1+u^2} = \arctan(t^2).$$
