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.