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 $.


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*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.

There is a short way: use equivalents to remove unnecessary details:


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*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 $[1,+\infty)$.

