self study - sum and product rules of probability - Cross
First, you apply the product rule in the integral. This yields $$p(x=1|\mathcal D) = \int_0^1 p(x=1,\mu|\mathcal D) \mathrm d \mu$$ This is basically the definition of the summation rule which integrates out $\mu$. A few comments on that . Note, that in the comments above you said that $\mu$ is in front of "|", but you wrote $p(x=1|\mu,\mathcal D)$.
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