An Integral Substitution for $\int_0^{1} dy \left(\frac{M^2(y)}{\mu^2}\right)^{-\epsilon}$

Question

I have integral (1) as a result from an advanced QFT problem.

$$ \tag{1} I= \frac{\alpha}{2\epsilon} \int_0^1 dy \left( \frac{M^2}{\mu^2} \right)^{-\epsilon} + \mathcal{O}(\epsilon) $$ where $\mu$ and $\epsilon$ are constants and $M$ is a function of $y$ as, $$M^2 \equiv q^2y(y-1) + m^2$$ I have trying to get it into the format of (2) as I have a solution for this equation, $$ \tag{2} \int_0^1 \frac{dy}{\sqrt{1-y}} (1-zy)^{-\epsilon} $$

Attempts:

1. Substitution $dy\to dM^2$ : This doesn't give the required square root
2. Substitution $dy\to dM$ : The square root ends up on the wrong side.
3. Logarithmic expansion of integrand: $\left( \frac{M^2}{\mu^2} \right)^{-\epsilon} \approx 1 - 2\epsilon\ln\tfrac{M}{\mu} + \mathcal{O}(\epsilon)$ which looses the required power of $\epsilon$
4. Logarithmic expansion in $M$ as $\left( \frac{M^2}{\mu^2} \right)^{-\epsilon} \approx \left(1 - \epsilon\ln M \right)(\mu^2 M)^{-\epsilon} + \mathcal{O}(\epsilon)$
5. I have also tried a substitution $dy \to d\ln M$ which failed to produce a square root