Autonomous differential equation Let $f: \Bbb R \to \Bbb R$ and $x_0 \in \Bbb R$, such that $f(x_0)> 0 $, and assume that $x(t)$ is the solution of $x'=f(x)$, such that $x(0)=x_0$. 
If $f(x) > 0$ then $x(t)$ is defined for all $t \geq 0$ if and only if 
$$
\int_{x_o}^\infty \frac {ds}{f(s)} = \infty.
$$
I haven't really tried much since I don't know where to start...
 A: Hint. Define
$$
F(x)=\int_{x_0}^x\frac{ds}{f(s)}.
$$
Then, $F$ is strictly increasing,  continuously differentiable and
$\lim_{x\to\infty}F(x)=\infty$. Also $F(x_0)=0$. In particular
$$
F :[x_0,\infty)\to [0,\infty)
$$
is 1-1, onto, with positive derivative. Indeed, $F'(x)=\dfrac{1}{f(x)}>0$. Hence $F$ possesses an inverse
$$
\varphi :[0,\infty)\to [x_0,\infty),
$$
which is also continuously differentiable with positive derivative. This means
$$
F\big(\varphi(t)\big)=t, \quad \text{for all $t\in [0,\infty)$}.
$$
This implies that
$$
1=\frac{d}{dt}F\big(\varphi(t)\big)=F'\big(\varphi(t)\big)\varphi'(t)
=\frac{1}{f\big(\varphi(t)\big)}\varphi'(t),
$$
and finally
$$
\varphi'(t)=F\big(\varphi(t)\big), \quad \text{for all $t\in [0,\infty)$},
$$
and also $\varphi(0)=F^{-1}(0)=x_0$.
Note. If $\displaystyle\int_{x_0}^\infty\frac{ds}{f(s)}=T<0$, then the solution
$\varphi$ of
the IVP
$$
x'=f(x), \quad x(0)=x_0,
$$
would only be definable in $[0,T)$, and we would have $\lim_{t\to T}\varphi(t)=\infty$.
