I think the method would be something like this:

1/1+sinx+sin²x dx

We have 1+sinx+sin²x = (1-sinx) (1-²sinx)

Hence 1/1+sinx+sin²x = 1/(1-sinx) (1-²sinx) = A/(1-sinx) + B/(1-²sinx)

Comparing LHS and RHS we get
A+B = 1 and
A²+B = 0

Eqn 2 gives on multiplying by ². A+B = 0 or A+1 = A giving
A = 1/1-  and B = -/1-

Now the integral can be evaluated using the integral dx/a+bsinx which I've unfortunately forgotten how to do.I leave it for the students to finish.
 

PS: I should have said: "The rest is left as an exercise for  the student".