How can I test the convergence of series $ \int_{0}^{1} \frac{\ln (x)}{(1-x^6)(\arccos x)}\,dx $ I have found that this function is continuous from $(0,1)$, and the problematic points are $0$ and $1$.
For point $0$, I tried to compare it with $\log(x)$ and found that it converges.
For point $1$, I found the limit of $\cfrac {f(x)}{g(x)}$, whereas $f(x)$ is the integrand and $g(x) = \cfrac{(x-1)}{(1-x)(1-x)^{1/2}}$ as $x \to 1$ and also found that it converges.
So to conclude this integrand converges absolutely.
However, I still feel like it still needs improvement. I would appreciate any comment that contributes to the answer to this problem. Thank you
 A: Here is a careful treatment of this problem. Since $y=1-x^6$ and $y=\arccos(x)$ are both positive and decreasing functions on the domain $(0,0.5)$ we can say for any $0<x<0.5$ that $$0<\frac{1}{(1-x^6)\arccos(x)}<\frac{1}{(1-(0.5)^6)\arccos(.5)}=\frac{64}{21 \pi}$$ Since $y= -\ln(x)$ is positive on $(0,0.5)$ we get $0<-\frac{\ln(x)}{(1-x^6)\arccos(x)}<-\frac{64\ln(x)}{21\pi}$ for any $0<x<0.5$. Because $\int_0^{0.5}-\frac{64\ln(x)\mathrm{d}x}{21\pi}$ converges to $\frac{32(\ln(2)+1)}{21\pi}$ we know by direct comparison that $\int_0^{0.5}\frac{-\ln(x)\mathrm{d}x}{(1-x^6)\arccos(x)}$ is necessarily convergent. On the other hand, because $-\frac{\ln(x)}{1-x^6}\rightarrow \frac{1}{6}$ as $x\rightarrow 1^{-}$ we can find $\delta_0>0$ such that $\Big|-\frac{\ln(x)}{1-x^6}-\frac{1}{6}\Big|<\frac{1}{10}$ for any $x\in (1-\delta_0,1)$ This implies $0<-\frac{\ln(x)}{1-x^6}<\frac{4}{15}$ for any $x$ belonging to the interval $(1-\delta,1)$ where $\delta=\min\{0.5,\delta_0\}$. Using the fact that $y=\arccos(x)$ is positive on $(1-\delta,1)$ we get $0<-\frac{\ln(x)}{(1-x^6)\arccos(x)}<\frac{4}{15\arccos(x)}$ on $(1-\delta,1).$  The improper integral $\int_{1-\delta}^1\frac{4\mathrm{d}x}{14\arccos(x)}$ converges by limit comparison test to $\int_{1-\delta}^{1}\frac{\mathrm{d}x}{\sqrt{1-x}}$ hence so must $\int_{1-\delta}^1-\frac{\ln(x)\mathrm{d}x}{(1-x^6)\arccos(x)}$ Since $\int_{0.5}^{1-\delta}-\frac{\ln(x)\mathrm{d}x}{(1-x^6)\arccos(x)}$ isn't event improper, we see that $\int_0^1-\frac{\ln(x)\mathrm{d}x}{(1-x^6)\arccos(x)}$ is convergent and so must be your integral.
