1. ## Incorrect Indefinite Integral Answers

I'm having difficulty with this question, I've been able to successfully get it as far as this (which is correct according to the solutions page) but I cannot get the actual stage of finding the indefinite integral part correct.
$\displaystyle \int(1-2e^{-x/2}+e^{x/2})dx$
Finding the indefinite integrals for each term at a time gives me:
$\displaystyle \int(1)dx=x+c$
$\displaystyle \int(-2e^{-x/2})dx=\int(-2)(e^{(-1)x/2})dx=(-2)\frac{1}{-1}e^{(-1)x/2}+c=2e^{-x/2}+c$
$\displaystyle \int(e^{x/2})dx=\int(e^{(1)x/2})dx=\frac{1}{1}e^{(1)x/2}+c=e^{x/2}+c$
Putting this all together gives me:
$\displaystyle \int(1-2e^{-x/2}+e^{x/2})dx=x+2e^{-x/2}+e^{x/2}+c$
But the correct solution is:
$\displaystyle \int(1-2e^{-x/2}+e^{x/2})dx=x+4e^{-x/2}+2e^{x/2}+c$
Can anybody see where I am making my mistakes?
Many thanks.

2. Third line, e^{x/2} is e^{x \times 1/2} and not what you got, so you need to times its integral by 2. Similarly for the line above except it's minus it.

3. Matt that's clear now, thank you very much.
Shayne.