How do you evaluate #\int ( e ^ { \frac { x } { 2} } - 4) ^ { 2} d x#?

1 Answer
Dec 16, 2017

#e^x-16e^(x/2)+16x+C#

Explanation:

#int(e^(x/2)-4)^2dx#

multiply the bracket out

#=int(e^(x/2)-4)(e^(x/2)-4)dx#

#=int((e^(x/2))^2-2(4)(e^(x/2))+16)dx#

#=int(e^x-8e^(x/2)+16)dx#

integrate term by term

#=e^x-16e^(x/2)+16x+C#