How to prove that the solution of $\frac{dy}{dx}=1+y^4$ with $y(x_0)=y_0$ cannot be extended to $\infty$ and $-\infty$ A friend of mine gave me a proof, but I didn't get it.
If the solution can be extended to +∞
$$ \frac{dy}{1+y^4} = dx $$
Integrate $x$ from $x_0$ to +∞
So $$+∞ = \int_{x_0}^\infty \frac{ dy(x)}{1+y^4 } (\alpha) $$
Let $y$ replace $x$
Then 
$$ \alpha = \int_{y_0}^y \frac{dy}{1+y^4} (+∞), $$ which is no greater than from $y_0$ to +∞
So the hypothesis is wrong
The solution cannot be extended to +∞
I don't think $dy$ change into $dy(x)$ is correct during the first integration process. Is this proof right?
Any suggestion will be appreciated.

Alternative argument: You have at all times $y'\ge 1$. Thus there is a time $x_1$ where $y(x_1)=1$ is reached. From that point on use $y'\ge y^4$ which can be easily solved as
$$
(y^{-3})'=-3y^{-4}y'\le -3\implies y(x)^{-3}-y(x_1)^{-3}\le -3(x-x_1)
\\ ~ \\
\implies y(x)\ge \frac1{\sqrt[3\,]{1-3(x-x_1)}}
$$
so that at some time before $x=x_1+\frac1{3}$ the solution diverges to infinity.
