The correct option is A
ex(2x3−6x2+7x+3)+c
∫ex(2x3−5x+10)dx
2x3−5x+10=(ax3+bx2+cx+d)+(3ax2+2bx+c)
a=2;b+6=0b=−6;c+2b=−5c=7;d+c=10d=3
2x3−5x+10=(2x3−6x2+7x+3)+(6x2−12x+7)
∫ex(2x3−5x+10)dx=∫ex[(2x3−6x2+7x+3)+(6x2−12x+7)]dx
It is in the form of ∫ex(f(x))+f(x)]dx=exf(x)+c
=ex(2x3−6x2+7x+3)+c