Q- the value of integration from (0 to -2)

(x(cube)+3x(sq.)+3+(x+1) cos(x+1) dx is

I=x^3+3x^2+3+(x+1)cos(x+1)

 ={x^3+1+3x(x+1)} +2-3x+(x+1)cos(x+1)          adding and subtracting -3x

 ={(x+1)^3)}+ 5 -3(x+1) +(x+1)cos(x+1)         

now put x+1=t

           so dx=dt

now becomes

              I=(t^3 -3t+tcost+5)lim +1to-1                   (t^3,t,tcost all are odd functions so integral will be 0)

              I=5(1+1)

                =10