Time in movement for forces as functions of position Let there be an object in rest of $1kg$ mass at $x=0$ and a force acted upon it which can be described by the equation
$F(x) = \frac{1}{(1-2x)^2}$ 
with $x$ belonging in $[0,1/2]$. I want to know in how much time it will move by $1/2 m$.
sο what I need is a way to connect  $x$ with $t$ (where $t$ is time).
neuton's second law of motion states that $F=ma$ where $m$ is the mass of the object and $a$ its acceleration 
so $F=ma=1a=a$ but $a= \frac{du}{dt}= \frac{du}{dx} \frac{dx}{dt}= \frac{du}{dx} u$ so $\frac{1}{(1-2x)^2} = \frac{du}{dx} u <=> \frac{dx}{(1-2x)^2} = u du <=> \frac{1}{2} \frac{1}{1-2x} = \frac{1}{2} u^2 <=> \frac{1}{1-2x} = u^2$

Should be
$$\int_0^x \frac{dx}{(1-2x)^2}=\int_0^u udu$$
$$\frac{1}{2}\left[\frac{1}{1-2x}\right]_0^x=\frac{u^2}{2}$$
$$u^2=\frac{1}{1-2x}-1$$
So
$$\frac{1}{(1-2x)^2}=(u^2+1)^2$$
Then
$$\frac{du}{dt}=\frac{1}{(1-2x)^2}=(u^2+1)^2$$
$$\int_0^u\frac{du}{(u^2+1)^2}=\int_0^t dt$$
By substitution with $u=\tan\theta$, we have
$$t=\frac{1}{2}\left[\arctan u + \frac{u}{1+u^2}\right]$$
Now, when $x\to 1/2$, $u\to+\infty$
Time taken is $\pi/4$.
