It is the integration-by-parts formula, although demonstrated in a quite confusing way. Especially, it is unclear which function $\int$ is applied to. Using parentheses to emphasize the scope of $\int$ for each instance, we may instead write
$$\int(fg) = \left(\int f\right) g - \int \left(\left( \int f \right) g'\right). $$
It is still an unconventional way of demonstrating the IbP formula. So let me revert it to the good old way. Let $F $ denote an anti-derivative of $f$, i.e., $F' = f$. Then
$$ \int f(x)g(x) \, \mathrm{d}x = F(x)g(x) - \int F(x)g'(x) \, \mathrm{d}x. $$
Alternatively, it may help to adopt programmer-style notation for this. Write $\mathtt{Integrate}[f]$ for any ani-derivative of $f$ and $\mathtt{D}[g]$ for the derivative $g'$. Then the handwriting translates to
$$ \mathtt{Integrate}[f \cdot g] = \mathtt{Integrate}[f] \cdot g - \mathtt{Integrate}[\mathtt{Integrate}[f] \cdot \mathtt{D}[g]]. $$