Evaluating the improper integral $\int_0^1 \frac{\log (x \sqrt{x})}{\sqrt{x}} \,dx$ I am supposed to solve this integral but i have no idea how:

$$\int_0^1 \frac{\log (x \sqrt{x})}{\sqrt{x}} \,dx$$

Since one limit is $0$ it will be divided by zero.
Can someone please explain this to me (I really want to understand) and guide me through step by step. Thanks! :)

Hint. You know that a primitive of $\log u$ is $u\log u -u$, then using the change of variable $\displaystyle u=\sqrt{x}$, $\displaystyle du=\frac 1{2\sqrt{x}}dx$, you obtain
$$
\int_0^1 \frac{\log (x \sqrt{x})}{\sqrt{x}} \,dx= 2 \int_0^1 \log (u^3) \,du=6\times[u\log u -u]_0^1=-6.
$$
