My textbook gives the RG equation: $$ \frac{d \bar{m}(\mu)}{d \ln \mu}=\gamma_{m} \bar{m}(\mu) ; \quad \gamma_{m}=-\frac{3 \alpha}{2 \pi} $$ And then says this is easy to solve and the solution is: $$ \begin{aligned} \bar{m}(\mu) &=\bar{m}\left(\mu_{0}\right) \exp \left[\int_{\ln \mu_{0}}^{\ln \mu} d \ln \mu^{\prime} \gamma_{m}\right] \\ &=\bar{m}\left(\mu_{0}\right) \exp \left[\gamma_{m} \ln \frac{\mu}{\mu_{0}}\right]=\bar{m}\left(\mu_{0}\right)\left(\frac{\mu}{\mu_{0}}\right)^{\gamma_{m}} \end{aligned} $$ Though I can't see how they make this jump to solve the equation, and there seems to be quite little about solving RG equations online. Please could someone break down how they have got to this solution?
EDIT: so I have got as far as: $$ \bar{m}(\mu) = \gamma_{m} \int \bar{m}(\mu) d\ln{\mu} $$ I am not sure how they got the form in the first line given from where I am currently.