here is my problem and what I have done so far.

\(\displaystyle \frac{dx}{dt} = x(1 - x)\)

\(\displaystyle \frac{dt}{dx} = \frac{1}{x(1 - x)}\).

Now using the method of partial fractions:

\(\displaystyle \frac{A}{x} + \frac{B}{1 - x} = \frac{1}{x(1 - x)}\)

\(\displaystyle \frac{A(1 - x) + Bx}{x(1 - x)} = \frac{1}{x(1 - x)}\)

\(\displaystyle A(1 - x) + Bx = 1\)

\(\displaystyle A - Ax + Bx = 1\)

\(\displaystyle A + (B - A)x = 1 + 0x\).

Therefore \(\displaystyle A = 1\) and \(\displaystyle B - A = 0\), so \(\displaystyle B = 1\).

Thus \(\displaystyle \frac{1}{x(1 - x)} = \frac{1}{x} + \frac{1}{1 - x}\).

Back to the DE:

\(\displaystyle \frac{dt}{dx} = \frac{1}{x(1 - x)}\)

\(\displaystyle \frac{dt}{dx} = \frac{1}{x}+ \frac{1}{1 - x}\)

\(\displaystyle t = \int{\left(\frac{1}{x} + \frac{1}{1 - x}\right)\,dx}\)

\(\displaystyle t = \ln{|x|} - \ln{|1 - x|} + C\)

\(\displaystyle t = \ln{\left|\frac{x}{1 - x}\right|} + C\)

\(\displaystyle t = \ln{\left|\frac{x - 1 + 1}{1 - x}\right|} + C\)

\(\displaystyle t = \ln{\left|\frac{-(1 - x)}{1 - x} + \frac{1}{1 - x}\right|} + C\)

\(\displaystyle t = \ln{\left|-1 + \frac{1}{1 - x}\right|} + C\)

\(\displaystyle t - C = \ln{\left|-1 + \frac{1}{1 - x}\right|}\)

\(\displaystyle e^{t - C} = \left|-1 + \frac{1}{1 - x}\right|\)

\(\displaystyle e^{-C}e^t = \left|-1 + \frac{1}{1 - x}\right|\)

\(\displaystyle \pm e^{-C}e^t = -1 + \frac{1}{1 - x}\)

\(\displaystyle A\,e^t = -1 + \frac{1}{1 - x}\), where \(\displaystyle A = \pm e^{-C}\)

\(\displaystyle A\,e^t + 1 = \frac{1}{1 - x}\)

\(\displaystyle \frac{1}{A\,e^t + 1} = 1 - x\)

\(\displaystyle x = 1 - \frac{1}{A\,e^t + 1}\)

\(\displaystyle x = \frac{A\,e^t + 1 - 1}{A\,e^t + 1}\)

\(\displaystyle x = \frac{A\,e^t}{A\,e^t + 1}\).