How do I write:
$$\int {dx\over \sqrt x\log x}$$
In terms of $u$:
$$\int {du \over \log u} \quad u = \sqrt x$$
I'm given the substitution for $u$, but I cannot reduce it further than $\int {2du\over \log x}$. How would I rewrite fully in terms of $u$?
$$u = \sqrt x \quad du = {1\over 2 \sqrt x}dx \quad dx = 2\sqrt x du$$
Thus, rewriting in terms of $u$:
$$\require{cancel} \begin{align} \int {dx \over \sqrt x \log x} &= \int {1\over \cancel{\sqrt x} \log x} 2\cancel{\sqrt x} du \\ &= \int {2\over \log x} du \\ \end{align}$$
Now, you can do some rewriting:
$$\begin{align} \int {2\over \log x} du &= \int {1\over {1\over 2} \log x} du \\ &= \int {1\over \log x^{\frac 12}} du \\ &= \int {1\over \log \sqrt x} du \\ &= \int {1\over \log u} du \\ \end{align}$$
Remember the log property $\log\left(a^b\right) = b\log(a)$. That's how you rewrite $\frac 12 \log x$ as $\log \sqrt x$, and thus $\log u$.