# From integral to $\Theta\left(\sqrt{n}\left(\log{n} +1\right)\right)$

Why is $$T_2(n) \in \Theta\left(n^{1/2}\left(1 + \int_1^{n}\frac{\sqrt{u}}{u^{1/2+1}}du\right)\right) \implies \Theta\left(\sqrt{n}\left(\log{n} +1\right)\right)$$

I know that $$n^{\frac{1}{2}} = \sqrt{n}$$, but how does one get from the integral to log $$n$$ + 1?

And I know that $$\Theta\left(\sqrt{n}\left(\log{n} +1\right)\right)$$ = $$\Theta(\sqrt{n} \log{n})$$

It looks like the term inside the integral equals $$\frac{1}{u}$$ as $$u^{1/2}=\sqrt{u}$$.
Hence it follows because $$\log(u)$$ is the anti-derivative of $$\frac{1}{u}$$.