how to integrate?

Question

e^ {[x-1]/2}

Answer 1

INT e^ [(x-1)/2] dx

let u = (x-1)/2. du = 1/2 dx. 2du = dx.

INT 2e^u du = 2 INT e^u du = 2e^u + C = 2e^[(x-1)/2] + C

Answer 2

Huh?? It's a simple exponential integral so 2e^((x-1)/2) + C
(Unless you typo'd it ☺)

Doug