check the convergence of the improper integral$\int_{0}^{1}\frac{x^{p-1}+x^{-p}}{1+x}\,dx$ How to check the convergence of the improper integral$$\int_{0}^{1}\frac{x^{p-1}+x^{-p}}{1+x}\,dx$$
I can only check that the integral is divergent for $p\geq1$, help for the cases when $p<1$.
Thanks!
 A: Hint: If $p\lt 1$ the $x^{-p}$ part doesn't matter. Throw it away.
Edit: Since you are confident about the divergence if $p\ge 1$, we suppose that $p\lt 1$. To be formal, we want to examine the behaviour of 
$$\int_\epsilon^1 \frac{x^{p-1}+x^{-p}}{1+x}\,dx$$
as $\epsilon$ approaches $0$ from the right. So we want to look into the existence/nonexistence of 
$$\lim_{\epsilon\to 0^+}\left(  \frac{x^{p-1}}{1+x}+ \frac{x^{-p}}{1+x}  \right)\,dx.$$
Since $p\lt 1$, $\lim_{\epsilon\to 0^+}\int_\epsilon^1 \frac{x^{-p}}{1+x}  \,dx$ exists. For note that $1+x\ge 1$ on our interval of integration.
So we need to determine whether $\lim_{\epsilon\to 0^+}\int_\epsilon^1 \frac{x^{p-1}}{1+x}\,dx$ exists. Note that $\frac{1}{2}\le \frac{1}{1+x}\le 1$ on our interval, so our problem is equivalent to determining whether $\lim_{\epsilon\to 0^+} \int_\epsilon^1 x^{p-1}\,dx$ exists. 
The answer is standard. Rewrite the function as $\frac{1}{x^{1-p}}$. The limit exists if $1-p\lt 1$,  and doesn't if $1-p\ge 1$. 
Remark: The derivation has been quite formal. One can state things quite a bit more informally. For $p\lt 1$, the problem comes down to the existence/nonexistence of $\int_0^1 \frac{dx}{x^{1-p}}$. 
A: My idea: substitute first
$$u:=\frac{1}{x}\implies dx=-\frac{du}{u^2}\implies$$
$$\int\limits_0^1\frac{x^{p-1}+x^{-p}}{1+x}dx=\int\limits_1^\infty \frac{u^{-p+1}-u^p}{1+u^{-1}}\frac{du}{u^2}=\int\limits_1^\infty \frac{u^{2p-1}+1}{u^{p+1}+u^{p}}du$$
$$\int\limits_1^\infty\frac{u^{2p-1}+1}{u^{p+1}+u^{p}}du\le\int\limits_1^\infty\frac{2u^{2p-1}}{u^{p+1}}du=\frac{2}{p-1}\;\lim_{b\to\infty}\;\left(b^{p-1}-1\right)=\frac{2}{1-p}$$
since $\,p<1\implies p-1<0\,$
