Ответ: ln(tg(x/2))+2/(1+tg(x/2) )+c

Объяснение:

Преобразуем знаменатель подынтегральной функции:

sin(x)*(1+sin(x) ) = 2*sin(x/2)*cos(x/2)* (sin(x/2)+cos(x/2) )^2=

=2*cos^4(x/2)*tg(x/2) *(1+tg(x/2))^2

dx/(sin(x) *(1+sin(x) ))=  dx/(cos^2(x/2) )/2*cos^2(x/2)*tg(x/2)*(1+tg(x/2))^2

cos^2(x/2)+sin^2(x/2)=1

1+tg^2(x/2)=1/cos^2(x/2)

int( (1+tg^2(x/2))*d(tg(x/2))/ ( tg(x/2)*(1+tg(x/2))^2) )

tg(x/2)=t

int( (1+t^2)*dt/t*(1+t)^2)

(1+t^2)/(t*(t+1)^2)  = ( (t+1)^2-2t)/(t*(t+1)^2) =1/t  -2/(1+t)^2

int( (1+t^2)*dt/t*(1+t)^2)=int( (1/t -2/(1+t)^2)dt)= ln(t)+2/(1+t)+c=                       ln(tg(x/2))+2/(1+tg(x/2) )+c