Determine all values of $a\geq0$ such that the improper integral $$\int^\infty_0\frac{\ln(1+x+x^a)}{x\sqrt{x}}$$ is convergent.
Since $D(\ln(1+x+x^a))=\frac{1+ax^{a-1}}{1+x+x^a}$ we have that $\lim_{x\rightarrow 0}\frac{\ln(1+x+x^a)}{x}=1$ by LH rule if $a\geq 1$ thus one can easily show that $\int^1_0\frac{\ln(1+x+x^a)}{x\sqrt{x}}$ is convergent if $a\geq 1$.
Since $\lim_{x\rightarrow\infty}\frac{\ln(1+x+x^a)}{x^{1,5}}/\frac{1}{x^{1,2}}=0$ we have that $\int^\infty_1\frac{\ln(1+x+x^a)}{x\sqrt{x}}$ is convergent for all $a$.
The answer should be $a> 1/2$ so how does one show that $\int^1_0\frac{\ln(1+x+x^a)}{x\sqrt{x}}$ is convergent for $1/2< a<1$?
Note that, if $a<1$, near $x=0$ you can use the Taylor expansion of logarithm: $$\frac{\log (1+x+x^a)}{x\sqrt x}\sim \frac{x^a}{x\sqrt x}=\frac{1}{x^{1+1/2-a}} $$ Using the fact that $\int_0^1 \frac{dx}{x^p}$ is convergent iff $p<1$, you get the condition $1+1/2-a<1$, namely $a>1/2$.
Note that you have to see also the behaviour as $x\to +\infty$ when studying the case $1/2<a<1$.
Edit
Recall that $\log(1+t)\sim t$, in this case $t=x+x^a$, so you have: $\log(1+x+x^a)\sim x+x^a$. Since $0<a<1$, and you want to see the expansion around $x=0$, small powers of $x$ dominate, namely $x+x^a\sim x^a$. You can see this also in this way: $x+x^a=x^a(x^{a-1}+1)$ and $x^{a-1}+1 \to 1$ as $x\to 0$ hence the dominant term is $x^a$.