So I have the definite integral:
$$\int_0^{\frac{uy}{(1-u)}} x^{(k/2)-1} e^{-(x+y/2)} \,dx$$
I have the answer, which is: $$(\frac{uy}{(1-u)})^{(k/2)-1} e^{-(\frac{\frac{uy}{(1-u)}+y}2)} (\frac y{{(1-u)}^2}) \,$$
but not sure where the: $$(\frac y{{(1-u)}^2})$$
is coming from. Any explanation would be much appreciated.
I've added a pic of the original problem, hope that clarifies the issue.
