For what values of $x$ is the expression $\frac{\log(x^2-4x-5+2x)}{\sqrt{x-1}}$ defined? Express your answer in interval notation.
First, we need to determine the values of $x$ that make the numerator and denominator defined.
The numerator is defined as long as $x^2-4x-5+2x$ is positive. Simplifying, we have $x^2-2x-5$. Factoring, we obtain $(x-1)(x+5)$. Therefore, the numerator is defined when $x>1$ or $x<-5$.
The denominator is defined as long as $x-1$ is positive. Therefore, the denominator is defined when $x>1$.
Putting it all together, the expression is defined when $x>1$. Therefore, the answer is $\boxed{(1, \infty)}$.
To find the values of $x$ for which the expression is defined, we need to consider the domains of the logarithm function and the square root function.
The logarithm function is defined only for positive values inside the logarithm. Therefore, we must have $x^2 - 4x - 5 + 2x > 0$.
Simplifying this inequality gives $x^2 - 2x - 5 > 0$.
Now, let's solve this quadratic inequality by finding its roots:
$x^2 - 2x - 5 = 0$
Using the quadratic formula, we have:
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-5)}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 20}}{2}$
$x = \frac{2 \pm \sqrt{24}}{2}$
$x = \frac{2 \pm 2\sqrt{6}}{2}$
$x = 1 \pm \sqrt{6}$
So the roots are $x = 1 + \sqrt{6}$ and $x = 1 - \sqrt{6}$.
To determine the sign of the inequality $x^2 - 2x - 5 > 0$, we can create a sign chart and test different regions:
\[
\begin{array}{|c|c|c|c|c|}
\hline
& x < 1 - \sqrt{6} & 1 - \sqrt{6} < x < 1 + \sqrt{6} & x > 1 + \sqrt{6} \\
\hline
x^2 - 2x - 5 & - & + & - \\
\hline
\end{array}
\]
From this sign chart, we can see that the inequality is true for $x < 1 - \sqrt{6}$ and $x > 1 + \sqrt{6}$.
Now let's consider the square root function. The expression is defined as long as the radicand $(x - 1)$ is non-negative. So we require $x - 1 \geq 0$, which means $x \geq 1$.
Therefore, combining the results, we find that the expression $\frac{\log(x^2-4x-5+2x)}{\sqrt{x-1}}$ is defined for $x \in \boxed{[1 + \sqrt{6}, +\infty)}$.