Show that this function is not square-integrable. Let $f(x):=(1-x^2)^{\frac{m}{2}} \int_0^x \frac{dt}{(1-t^2)^{m+1}}$ be a function on $(-1,1)$. Then I would like to show that the asymptotic of $f$ is such that for $m \in \{1,2,3,...\}$ the function $f$ is not square-integrable at $ \pm 1.$

You need only look at the case $x=1$. Put $\displaystyle F_m(x)=\int_0^x \frac{dt}{(1-t^2)^{m+1}}$. Then we compute that $F_0(x)=\frac{1}{2}\log (\frac{1+x}{1-x})$. Integrating $F_m$ by parts, we get that
$$(2m+2)F_{m+1}(x)=(2m+1)F_m(x)+\frac{x}{(1-x^2)^{m+1}}$$
Now it is easy to show that $(1-x^2)F_0(x)\to 0$ if $x\to 1$, and hence $(1-x^2)F_1(x)\to 1/2$ if $x\to 1$. Now we prove by induction that for $m\geq 1$, $(1-x²)^mF_m(x)\to \frac{1}{2m}$, and it is easy to finish. 
