Здравствуйте, Андрей!
4.
(x^4+3x^3+3x^2-5)/(x^3+3x^2+3x+1)=(x^4+3x^3+3x^2+x-x-5)/(x+1)^3=(x(x+1)^3-(x+5))/(x+1)^3=
=x-(x+5)/(x+1)^3=x-(x+1)/(x+1)^3-4/(x+1)^3=x-1/(x+1)^2-4/(x+1)^3;
[$8747$]((x^4+3x^3+3x^2-5)/(x^3+3x^2+3x+1))dx=[$8747$]xdx-[$8747$](1/(x+1)^2)dx-4[$8747$](1/(x+1)^3)dx=
=[$8747$]xdx-[$8747$](1/(x+1)^2)d(x+1)-4[$8747$](1/(x+1)^3)d(x+1)=
=x^2/2+1/(x+1)+2/((x+1)^2)+C.
5.
[$8747$]cos(x)cos(2x)cos(3x)dx=(1/2)[$8747$](cos(x)+cos(3x))cos(3x)dx=
=(1/2)[$8747$](cos(x)cos(3x)+cos^2(3x))dx=(1/4)[$8747$]((cos(2x)+cos(4x))+(1+cos(6x))dx=
=(sin(2x))/8+(sin(4x))/16+x/4+(sin(6x))/24+C=x/4+(sin(2x))/8+(sin(4x))/16+(sin(6x))/24+C.
С уважением.
Об авторе:
Facta loquuntur.