$f(x)={{\left| 1-x \right|}^{-0.5}}\exp (-{{\left| 1-x \right|}^{0.5}})$ for $x>0$
I was thinking to do a u-sub, $u={{\left| 1-x \right|}^{0.5}}$
but what would your $du$ be?
should I consider the negative sign inside the absolute value?
$f(x)={{\left| 1-x \right|}^{-0.5}}\exp (-{{\left| 1-x \right|}^{0.5}})$ for $x>0$
I was thinking to do a u-sub, $u={{\left| 1-x \right|}^{0.5}}$
but what would your $du$ be?
should I consider the negative sign inside the absolute value?
If $u=\sqrt{|1-x|}$, then in fact $u=\begin{cases} \sqrt{1-x},&\text{if }x\le 1\\\\ \sqrt{x-1},&\text{if }x>1\;, \end{cases}\tag{1}$
since by definition $|1-x|=\begin{cases} 1-x,&\text{if }1-x\ge 0\\\\ -(1-x)=x-1,&\text{if }1-x<0\;. \end{cases}$
Thus, $du=\begin{cases} -\frac12(1-x)^{-1/2}dx,&\text{if }x<1\\\\ \frac12(x-1)^{-1/2}dx,&\text{if }x>1\;. \end{cases}\tag{2}$
Thus, you’ll need to split the integral in two, one for $x<1$ and one for $x>1$. (Why did I change $x\le 1$ in $(1)$ to $x<1$ in $(2)$?)