Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ 
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx 
$$ 


First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $.


*

*For $\int_0^{1}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I prove $\ln(1+x^2)< x^2$, using that I can prove $\int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ converges, but it take too long so is there shorter way to do this problem?  

*For $\int_1^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I have no idea how to do this problem.
 A: There is a short way: use equivalents to remove unnecessary details:


*

*For $\displaystyle \int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,\mathrm d x$, the integrand is equivalent near $0$ to
$$\frac{x^2}{\sqrt{2x^5}}=\frac1{\sqrt{2x}},$$
and the integral of the latter converges on $(0,1]$.

*For $\displaystyle \int_1^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,\mathrm dx $, the integrand is equivalent, when $x\to\infty$, to
$$\frac{\ln (x^2)}{\sqrt {x^6}}=\frac{2\ln x}{x^3}= \frac{\ln x}{x}\frac1{x^2}=o\Bigl(\frac1{x^2}\Bigr),$$
and the integral of $1/x^2$ converges on $[1,+\infty)$.

A: One of the most powerful tools for deciding convergence of improper integrals (infinite series in the discrete case) is the comparison test. However, making strict comparisons can be tedious and not necessary. The limit version of the comparison test is usually much easier to apply. 
Suppose one wants to decide whether $\int_0^1f(x)\,dx<\infty$, where $f$ is some nonnegative continuous function here. But $f$ is of some "ugly" expression that is hard to integrate or even hard to compare directly with certain known function on the whole interval $(0,1)$. However, if one has the following two easy to get facts for some positive function $g$:


*

*$\displaystyle\lim_{x\to 0+}\frac{f(x)}{g(x)}=C$ with $0\le C<\infty$;

*$\int_0^1g(x)\,dx<\infty$.


Then we can tell that $\int_0^1f(x)\,dx<\infty$. We have similar arguments for analyzing $\int_1^\infty f(x)\,dx$. 
See for instance this set of lecture notes. Such arguments would justify the asymptotic analysis in Bernard's answer. 

Since
$$
\lim_{x\to 0} \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\cdot\sqrt{x}
= \lim_{x\to 0} \frac{\ln (1 + x^2)}{x^2}\cdot
\frac{x^2\sqrt{x}}{\sqrt {2x^5 + x^6}}=\frac{1}{\sqrt{2}}
$$
and
$
\int_0^1\frac{1}{\sqrt{x}}\,dx<\infty,
$
we can conclude by the limiting comparison test that 
$$
\int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx<\infty.
$$
On the other hand, since
$$
\lim_{x\to \infty} \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\cdot x^2
=\lim_{x\to \infty}\frac{\ln (1 + x^2)}{x}\cdot\frac{x^3}{\sqrt {2x^5 + x^6}}
= 0 
$$
 and $
\int_1^\infty \frac{1}{x^2}\,dx<\infty,
$
 we have
$$
\int_1^\infty\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx<\infty.
$$

† In the argument above, one can use L'Hopital rule to find the following two limits. 
$$
\lim_{x\to 0+}\frac{\ln(1+x^2)}{x^2}=\lim_{x\to 0+}\frac{\ln(1+x)}{x}=1,\quad
\lim_{x\to\infty}\frac{\ln(1+x^2)}{x}=0
$$
A: Your approach on $[0,1]$ seems fine. For the other part, you can verify that:
$$\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}} \le \frac{\ln (1 + x^2)}{\sqrt {x^6}}=\frac{\ln (1 + x^2)}{x^3} \quad \mbox{and} \quad \int_1^{\infty}\frac{\ln (1 + x^2)}{x^3}\,\mbox{d}x = \ln 2$$
A: Just for the fun of it.
The convergence issues having been clearly explained and one answer for the value showing that a CAS is able to find a (nasty) antiderivative, the result is in fact
$$\int_0^{\infty}\frac{\log (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx=\frac{1}{3} \log \left(\frac{1}{4} \left(-1+i+\sqrt{-1-2 i}\right)^{-\sqrt{4-2 i}}\,\,
   \left(-1-i+\sqrt{-1+2 i}\right)^{-\sqrt{4+2 i}}\right)$$
A: For small $x$, $$\frac{\log(1+x^2)}{x^2\sqrt{2x+x^2}}=\frac1{\sqrt{2x}}+o(x^{-1/2}).$$
For large $x$, $$\frac{\log(1+x^2)}{x^2\sqrt{2x+x^2}}=\frac{\log(x)}{x^3}+o\left(\frac{\log(x)}{x^3}\right)=o\left(x^{-2}\right).$$
In both cases, the integrals of the asymptotic expressions do converge.
