Evaluating the improper integral $\int_0^1 \frac{\log (x \sqrt{x})}{\sqrt{x}} \,dx$

Question

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! :)

Answer 1

Use the transformation $x=z^2$, to get $$I=6\int_{0}^1 \ln z dz $$ and then use integration by parts to obtain $I=-6$.

Answer 2

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 $$ \int_0^1 \frac{\log (x \sqrt{x})}{\sqrt{x}} \,dx= 2 \int_0^1 \log (u^3) \,du=6\times[u\log u -u]_0^1=-6. $$

Answer 3

There are two things I would do first looking at this integral.

The first is resolve the $x\sqrt{x}$ into $x^\frac{3}{2}$, then use the log laws to write $\frac{3}{2}\log(x)$.

This gives you a much nicer log expression, which then makes integration by parts a much easier process.