General solution to $ \frac{dx}{dt} = x^a(1-x)^b$ My question is about the general solution for the following differential equation: 
$$ \frac{dx}{dt}  = x^a(1-x)^b,\quad a,b\gt 0~~~~~~~~~~~~~~~(1)~. $$
Obviously, if $a=b=1$ then (1) reduces to 
$$ \frac{dx}{dt}  = x(1-x) $$ which has as solution $$ x(t) = \frac{1}{1 + A e^{-t}}\,,$$ for some constant, $A$. In fact, for $a,b$ positive integers, a solution can be obtained by using method of separation of variables and partial fractions. I want to be able to find a solution that considers all cases and which would obviously include the special cases when $a$ and $b$ are positive integers.
 A: Well, your differential equation is separable:
$$\int \frac{1}{x^a(1-x)^b}~dx=\int dt \tag{1}$$
The left hand side cannot be integrated in terms of elementary functions.

However, one can directly apply the definition of the incomplete beta function:
$$\operatorname*{B}(x;\,m,n) = \int_0^x t^{m-1}\,(1-t)^{n-1}\,dt \tag{2}$$

If we put $(1)$ into a similar form:
$$\int x^{-a}(1-x)^{-b}~dx=t+c_1 \tag{3}$$
Hence, after applying $(2)$, the integral on the left hand side is evaluated as:
$$\int x^{-a}(1-x)^{-b}~dx=\operatorname*{B}(x,1-a,1-b)+c_2$$
Therefore, the general solution in implicit form is simply:
$$\bbox[5px,border:2px solid #C0A000]{\operatorname*{B}(x;1-a,1-b)=t+C}$$
A: Well, we have:
$$\text{x}'\left(t\right)=\text{x}\left(t\right)^\text{a}\cdot\left(1-\text{x}\left(t\right)\right)^\text{b}\space\Longleftrightarrow\space\int\frac{\text{x}'\left(t\right)}{\text{x}\left(t\right)^\text{a}\cdot\left(1-\text{x}\left(t\right)\right)^\text{b}}\space\text{d}x=\int1\space\text{d}x\tag1$$
For the integrals:


*

*$$\int1\space\text{d}x=x+\text{C}_1\tag2$$

*Substitute $\text{u}=\text{x}\left(t\right)$:
$$\int\frac{\text{x}'\left(t\right)}{\text{x}\left(t\right)^\text{a}\cdot\left(1-\text{x}\left(t\right)\right)^\text{b}}\space\text{d}x=\int\frac{1}{\text{u}^\text{a}\cdot\left(1-\text{u}\right)^\text{b}}\space\text{d}\text{u}\tag3$$

