What to know about convergence of integrals According to the values of p>0 examine the convergence of the integral:
$$\int_0^{+\infty} \dfrac{\ln(1+2x^{3p})}{(x+x^2)^{4p}\arctan(x)^{1/2}}dx$$
I didn't find a good explanation about this kind of problems,so i will be glad if someone say a few words about it.
 A: 
Integral over $(1,\infty)$

$$g(x):=\dfrac{\ln(1+2x^{3p})}{(x+x^2)^{4p}\arctan(x)^{1/2}}\\
\int_1^\infty g(x)dx<\infty\iff\sum_{n=1}^\infty g(n)<\infty\qquad\text{by integral test}\\
$$
As $\lim_{n\to\infty}\arctan(n)={\pi\over2}>1$, for suitably large $n$ we have
$${1\over2^{4p}\sqrt{\pi/2}}{\color{red}{\ln2}+\color{blue}{3p\ln n}\over n^{8p}}={\ln(2n^{3p})\over(2n^2)^{4p}\sqrt{\pi/2}}<g(n)<{\ln(3n^{3p})\over(n^2)^{4p}}={\color{red}{\ln3}+\color{blue}{3p\ln n}\over n^{8p}}
$$
$$
\color{blue}{\sum_{n=1}^\infty{\ln n\over n^{8p}}<\infty\iff\sum_{n=1}^\infty{n\over 2^{(8p-1)n}}<\infty\qquad\text{by Cauchy condensation test}\\
8p>1\iff\sum_{n=1}^\infty{n\over 2^{(8p-1)n}}<\infty\qquad\text{by root test}}\\
\color{red}{\sum_{n=1}^\infty{1\over n^{8p}}<\infty\iff\sum_{n=1}^\infty{1\over 2^{(8p-1)n}}<\infty\qquad\text{by Cauchy condensation test}\\
8p>1\iff\sum_{n=1}^\infty{1\over 2^{(8p-1)n}}<\infty\qquad\text{by root test}}\\
\therefore 8p>1\iff \int_1^\infty g(x)dx<\infty
$$

Integral over $(0,1)$

$$f(x):={\ln(1+0.5x^{-3p})\over x^2(x^{-1}+x^{-2})^{4p}\sqrt{\arctan(x^{-1})}}\implies\int_{0}^1g(x)dx=\int_1^{\infty}f(x)dx$$
As $$\arctan(x)=\sum_{n=0}^\infty(-1)^n{x^{2n+1}\over{2n+1}}\\\ln(1+x)=\sum_{n=0}^\infty(-1)^n{x^{n+1}\over{n+1}}$$ for suitably large $n$ we have
$$\begin{align}
{0.5n^{-3p}-0.25n^{-6p}\over n^2(2n^{-1})^{4p}n^{-0.5}}<&f(n)<{0.5n^{-3p}\over n^{2-4p}(n^{-1}-n^{-3}/3)^{0.5}}\\
\implies{1\over2^{4p+2}}{2n^{-3p}-n^{-6p}\over n^{1.5-4p}}<&f(n)<{0.5n^{-3p}\over n^{2-4p}(n^{-1}-n^{-1}/3)^{0.5}}\\
\implies {1\over2^{4p+2}}{2n^{-3p}-n^{-3p}\over n^{1.5-4p}}<&f(n)<{0.5\sqrt{1.5}\over n^{1.5-p}}\\
\implies\qquad\quad {1\over2^{4p+2}}{1\over n^{1.5-p}}<&f(n)<{0.5\sqrt{1.5}\over n^{1.5-p}}
\end{align}\\
p<0.5\iff\sum_{n=1}^\infty f(n)<\infty\iff\int_1^{\infty}f(x)dx<\infty\iff\int_{0}^1g(x)dx<\infty$$

Conclusion

$$0.125<p<0.5\iff\int_{0}^\infty\dfrac{\ln(1+2x^{3p})}{(x+x^2)^{4p}\arctan(x)^{1/2}}dx<\infty$$

References

Integral test.  Cauchy Condensation test. Root test.
A: We are given that $p>0$.  We will now analyze convergence at both the lower and upper limits of integration.
Lower Limit
We have the following results from expanding terms for small $x$
$$\begin{align}
\frac{\log (1+2x^{3p})}{x^{4p}(1+x)^{4p}\arctan(x)^{1/2}}&=\frac{2x^{3p}+O(x^{6p})}{x^{4p+1/2}(1+O(x))}\\\\
&=2x^{-p-1/2}+O(x^{\min(2p-1/2,-p+1/2)})
\end{align}$$
So, provided that $p<1/2$, the lower limit poses no problem to the convergence of the integral.

Upper Limit
We have the following results from expanding terms for large $x$
$$\begin{align}
\frac{\log (1+2x^{3p})}{x^{4p}(1+x)^{4p}\arctan(x)^{1/2}}&=\frac{\log2x^{3p}+O(x^{-3p})}{x^{8p}(1+O(x^{-1}))(\pi/2+O(x^{-1}))}\\\\
&=x^{-8p}\log(2x^{3p})(1+O(x^{-1}))+O(x^{-11p})
\end{align}$$
So, provided that $p<1/8$, the upper limit poses no problem to the convergence of the integral.

Putting it all together, we have 
$$\bbox[5px,border:2px solid #C0A000]{\text{The integral converges for}\,\,1/8<p<1/2}.$$
