Convergence of Improper Integral $\int_0^1\frac{dx}{\ln(1+\sqrt{x})}$ I am struggling to understand what is the reasoning to get the following results:
Question
The integral
$$\int_0^1\frac{dx}{\ln(1+\sqrt{x})}$$
Converges.
My Attempt
I tried using the fact that for any two continuous function $f(x)<g(x)$, if $g(x)$ converges, $f(x)$ converges. Nonetheless, I could not find the right function to compare it to, since something like $\frac{1}{\ln{\sqrt{x}}}$ diverges and would not bound the integral. Any help on either the approach or finding a function is widely appreaciated.
 A: We have generally
$$x \geq \log(1+x)$$
but also for $x\in[0,a]$
$$\frac{\log(1+a)}{a}x \leq \log(1+x)$$
On $[0,1]$ then we have
$$\log 2 \cdot \sqrt{x} \leq \log(1+x) \implies \frac{1}{\sqrt{x}\log 2} \geq \frac{1}{\log(1+x)}$$
and
$$\int_0^1\frac{dx}{\log(1+\sqrt{x})} \leq \frac{1}{\log 2}\int_0^1\frac{dx}{\sqrt{x}} = \frac{2}{\log 2}$$
so the integral converges.
A: A possible approach could be based on the fact that the coefficients of the Taylor expansion of the integrand are alternating in sign. So, we can develop upper and lower bounds. A simple one could be
$$\frac{1}{\sqrt{x}}+\frac{1}{2}-\frac{\sqrt{x}}{12}<\frac{1}{\log \left(1+\sqrt{x}\right)} < \frac{1}{\sqrt{x}}+\frac{1}{2}$$
$$\frac{22}{9}<\int_0^1\frac{1}{\log \left(1+\sqrt{x}\right)}< \frac{5}{2}$$
A: Let $f(x)=\frac{1}{\ln (1+\sqrt{x})}$ 
Define a new function
$$g(x)=\frac{1}{\sqrt{x}}$$
Now ,
$$ \lim_{x\to 0^{+}} \frac{f(x)}{g(x)}=\lim_{x\to 0} \frac{\sqrt{x}}{\ln(1+\sqrt{x})} = 1  \neq 0, \infty $$
Hence by comparison test 
$\int_{0}^1 f(x) dx$ and $\int_{0}^1 g(x) dx$ converge or diverge together.

But note that
$$\int_{0}^1 \frac{1}{\sqrt{x}} = 2$$ which is convergent. Therefore the given integral is also convergent.
