Show that $\int_0^1\frac{1}{(x+1)(x+2)\sqrt {x(1-x)}} \,dx$ is convergent.
The points $0$ and $1$ are the only points of infinite discontinuities of $\frac{1}{(x+1)(x+2)\sqrt {x(1-x)}}.$ The integral is convergent is convergent if for a real number $c\in (0,1)$ the integrals $\int_0^c\frac{1}{(x+1)(x+2)\sqrt {x(1-x)}}\,dx$ and $\int_c^1\frac{1}{(x+1)(x+2)\sqrt {x(1-x)}}\,dx$ are convergent. Let $c=\frac 12.$ Now, we consider $I=\int_0^{\frac 12}\frac{1}{(x+1)(x+2)\sqrt {x(1-x)}}\,dx.$
Let $g(x)=\frac{1}{\sqrt {x(1-x)}}$ and $f(x)=\frac{1}{(x+1)(x+2)\sqrt {x(1-x)}}$ We note that, $\lim_{x\to 0+}\frac{f(x)}{g(x)}=\lim_{x\to 0+}\frac{1}{(x+1)(x+2)}=\frac 12\in \Bbb R^*.$
By Comparison Test (Limit Form), $\int_0^{\frac 12}f(x)dx$ and $\int_0^{\frac 12}g(x)dx$ converge or diverge together.
Let's examine the convergence of $\int_0^{\frac 12}g(x)dx.$
Firstly, $0$ is the only point of infinite discontinuity of $g(x).$ Let $\phi(\epsilon)=\lim_{\epsilon\to 0_+}\int_{\epsilon}^{\frac 12}g(x)\,dx.$ So, $$\lim_{\epsilon\to 0_+}\phi(\epsilon)=\lim_{\epsilon \to 0_+}\int_{\epsilon}^{\frac 12}\frac{1}{\sqrt{x(1-x)}}dx=I.$$
But $\int g(x)dx=\int \frac{1}{\sqrt{x-x^2}}\,dx=\int \frac{1}{\sqrt{x-x^2-\frac14+\frac14}}dx=\int\frac1{\sqrt{\frac 14-(x-\frac 12)^2}}\,dx=\log (\frac{x}{1-x}).$
Hence, $I=\lim_{\epsilon\to 0_+}\big [\log(\frac{x}{1-x})\big]_{\epsilon}^{\frac 12}=-\lim_{\epsilon\to 0_+}\log(\frac{\epsilon}{1-\epsilon}).$
So, we see that $I$ doesn't exist in $\Bbb R$ which means $\int_0^{\frac 12}g(x)dx$ is divergent which further implies $\int_0^{\frac 12}f(x)dx$ is divergent due to which we find that $\int_0^1\frac{1}{(x+1)(x+2)\sqrt {x(1-x)}}\,dx$ is divergent.
I got the opposite result as to what was required to show. I am not getting where I made the mistake. The proof looks fine to me. Any help regarding this apparent issue will be greatly appreciated.
