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).

Indeed, $d(x^2)=2xdx$. Then you can writes 
$$\int \frac{1}{1+x^4}\color{Blue}{2x dx} = \int \frac{1}{1+x^4}\color{Blue}{d(x^2)}$$ 
If $y=x^2$ then
$$\int \frac{1}{1+x^4}\color{Blue}{d(x^2)}= \int  \frac{1}{1+y^2}  dy = \arctan(y)+K= \arctan(x^2)+K,$$ $K$ constant.
