Please solve!!!?

Question

∫ √s+3 (s+1)^2 ds

Answer 1

█ next time don’t save on parentheses (-) and asterixes * to notify multiplication.
♠ Y(s)=∫√(s+3(s+1)^2) ds; y(s)= √(s +3s^2 +6s +3) =√(3s^2 +7s +3) =
=√((√3*s)^2 +2*√3*(3.5/√3)*s +(3.5/√3)^2 -(3.5/√3)^2 +3) =
=√((√3*s +3.5/√3)^2 -(3.5/√3)^2 +3);
now let a=√((3.5/√3)^2-3); Y(s)= a*√( ((√3*s +3.5/√3)/a)^2 -1 );
♣ let (√3*s +3.5/√3)/a =ch(t), d(cht) =sht*dt = (√3/a)*ds;
thus y(t)= a*√((cht)^2 -1)* (a*sht/√3)*dt =(a^2/√3) *(sht)^2 *dt=
=(a^2/√3)* 0.5*(ch(2t) -1)*dt; thus I =∫(ch(2t) -1)*dt =0.5sh(2t)-t =sht*cht-t;
♦ from t back to s variable: if x=cht then sht=√(x^2-1) and t=ln(x-√(x^2-1))
and I =x*√(x^2-1) +ln(x-√(x^2-1));
♥ Y(s)= (0.5/√3)*a^2 *I =(0.5/√3)*{(√3*s +3.5/√3)*√(3s^2 +7s +3) +((3.5/√3)^2-3) *ln(√3*s +3.5/√3 -√(3s^2 +7s +3))} +C; (poor whee!)
☺☺ unless anything is clear address me.

Answer 2

= 2/3 s^3/2 + (s+1)^3

Answer 3

dork u r a ******* nerd i bet your in college trying to figure something that "hard" out. idiot s=14