Consider $$\theta(x)=x+O(x\cdot\exp(-\delta^3(\log x)^{1/10}))$$ and the term $$\frac{\theta(x)}{\log x} - \int_2^x \theta(u)\frac{\mathrm{d}}{\mathrm{d}u}\Big(\frac{1}{\log u}\Big)\mathrm{d}u$$
What my textbook does now is, plugging in the error term for $\theta(x)$ and they get the result
$$\frac{x}{\log x} - \int_2^x u\frac{\mathrm{d}}{\mathrm{d}u}\Big(\frac{1}{\log u}\Big)\mathrm{d}u + O(x\cdot\exp(-\delta^3(\log x)^{1/10}))$$
However, I do not understand how we get this error term outside the integral. Consider $g(z):=x\cdot\exp(-\delta^3(\log x)^{1/10})$. If I simply plug the term in I would get sth like
$$\frac{\theta(x)}{\log x} - \int_2^x \theta(u)\frac{\mathrm{d}}{\mathrm{d}u}\Big(\frac{1}{\log u}\Big)\mathrm{d}u \\ = \frac{x}{\log x} + \frac{O(g(z))}{\log(x)} - \int_2^x u\frac{\mathrm{d}}{\mathrm{d}u}\Big(\frac{1}{\log u}\Big)\mathrm{d}u - \int_2^x O(g(u))\frac{\mathrm{d}}{\mathrm{d}u}\Big(\frac{1}{\log u}\Big)\mathrm{d}u$$
and unfortunately I do not see how I get the error term the textbook gets. I also think this is no legit notation, just writing $O(g(z))$ under the integral like this, but I know it is clear what I mean resp. where my problem is at. I'd appreciate any help how to handle this.
In essence, we are estimating
$$ R(x)=\int_2^x{g(u)\over u(\log u)^2}\mathrm du $$
It is not difficult to show that $g(u)u^{-1/2}$ is increasing, so we have
$$ R(x)\le{g(x)\over\sqrt x}\int_2^x{\mathrm du\over\sqrt u(\log u)^2}\ll g(x) $$