By substitution change 21–28 to integrals of rational functions

Solve $dy/dt=1-y^2$ by separating $\int dy/1-y^2=\int dt$. Then
$$\frac{1}{1-y^2}=\frac{1}{(1-y)(1+y)}=\frac{1/2}{1-y}+\frac{1/2}{1+y}$$
Integration gives $\frac{1}{2}\ln$ ____ $=t+C$. With $y_0=0$ the constant is $C=$ ____. Taking exponentials gives ____. The solution is $y=$ ____. This is the S-curve.

By substitution change 21–28 to integrals of rational functions. Problem 23 integrates $1/\sin \theta$ with no special trick.

This exercise builds on another exercise in its source book, which isn't shown here.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §7.4 Partial Fractions, #20