integral(x/√(3+x)) ???

∫(x/√(3+x))dx
Пусть u=x+3  ⇒  du=dx, a x=u-3
∫((u-3)/√u)du=∫√u-∫(3/√u)=∫u¹/²-3*∫u⁻¹/²=(2/3)*u³/²-3*2*u¹/²=
=(2/3)*(x+3)³/²-6*(x+3)¹/²=(2/3)*(x+3)¹/²*((x+3)-9)=(2/3)*√(x+3)*(x-6).