Integral from 0 to 1 (1-x)^{9 }
I did u substitution
u=1-x
du=-1dx
du/-1 =dx
get new intervals
1-1=0
1-0=1
-1 integral u^9 du
-1 u^{10}/10 interval 1 to 0
-1 (0^{10}/10 - 1^{10}/10)
=1/10
Am I correct?
You certainely noticed that $\displaystyle \frac1{(1-x)^9}$ is not defined for x = 1, that means the graph of $\displaystyle f(x)=\frac1{(1-x)^9}$ has a vertical asymptote x = 1. Therefore the definite integral doesn't exist.
EDIT: Sorry, I misunderstood your question completely!