_ -2 ^0 (x^3+3 x^2+3 x+3+(x+1) (x+1) ) d x is equals to

Step-by-step solution

Let aligned I & = _ -2 ^0 x^3+3 x^2+3 x+3+(x+1) (x+1) d x I & = _ -2 ^0 [(x+1)^3+2+(x+1) (x+1) d x . aligned Putting, x+1=t . Then d x=d t , we get aligned & I= _ -1 ^1 [t^3+2+t t ] d t & I= _ -1 ^1(2) u. =[2 t]_ -1 ^1=2[1-(-1)]=4 aligned As t^3+t t is an odd function.