x^2+4x+1=x^2+4x+4-3=(x+2)^2-3
= ∫ dx/((x+2)^2-3)[ замена х+2=u; dx=du]
= ∫ du/(u^2-3)=[ табличный интеграл]=
=(1/(2*sqrt(3))) *ln|(u-sqrt(3))/(u+sqrt(3))|+C=
=(1/(2*sqrt(3))) *ln|(x+2-sqrt(3))/(x+2+sqrt(3))|+C
О т в е т. (1/(2*sqrt(3))) *ln|(x+2-sqrt(3))/(x+2+sqrt(3))|+C