Non-linear ode solution with integral formula I am trying to figure out how the ode $\dot{x}=x^2, x(0)=c>1$ has the solution $x(t)=(\frac{1}{c}-t)^{-1}$ using the formula $x(t) = x_0+ \int^{t}_{t_0}f(s,x(s))ds$. I tried the substitution suggested in the comments $u=\frac{1}{x}$ but it doesnt give the correct form. I've also tried thinking about linearization but cant figure out how it fits into the integral formula?

Suppose we are given the initial value problem (IVP)
$$\dot{x}=f(t,x(t)),\quad x(0)=x_0\tag{1}$$
and the integral equation
$$x(t) = x_0+ \int^{t}_{0}f(s,x(s))\,ds\tag{2}$$
where $f$ is continuous in its domain. Then, $x(t)$ is a solution of the IVP $\iff$ $x(t)$ is a solution of the integral equation.
First, assume $x(t)$ is a solution to the integral equation. Then, 
$$\dot{x}=f(t,x(t)),\quad x(0)=x_0$$
It follows that
$$\frac{dx}{dt}=f(t,x(t))\implies\int_0^t dx=\int_0^t f(s,x(s))\,ds \implies x(t)-x(0)=\int_0^t f(s,x(s))\,ds $$
Next, assume that $x(t)$ is a solution to the integral equation
$$x(t) = x_0+ \int^{t}_{0}f(s,x(s))\,ds$$
Then, $f(s,x(s))$ is integrable from $0$ to $t$ for any $t$ in the domain. Thus, $\int_0^t f(s,x(s))\,ds$ is continuous in $t$. By $(2)$, we know that $x(t)$ is continuous. Moreover, since $f$ is continuous in its domain, $f(t,x(t))$ is continuous, from which it follows that the right-hand side of $(2)$ is differentiable, and so is $x(t)$. Differentiating both sides of $(2)$ we have $\dot{x}=f(t,x(t))$ where $x(0)=x_0$. Hence, $x(t)$ is a solution to the IVP $(1)$.
Given that $x(t)$ is a solution to both the IVP and the integral equation, we can work directly from $(1)$
$$\dot{x}=\frac{dx}{dt}=f(t,x(t))=x^2$$
which is a separable equation
$$\frac{1}{x^2}\,dx=dt$$
which forms
$$-x^{-1}+c_1=t \implies x(t)=\frac{1}{c_2-t}$$
the initial condition of $x(0)=c$ produces
$$c=\frac{1}{c_2}$$
therefore
$$x(t)=\frac{1}{\dfrac{1}{c}-t}=\left(\dfrac{1}{c}-t\right)^{-1}$$
