Improper integral $\int _{0+0}^{1-0}\frac{dx}{\left(4-3x\right)\sqrt{x-x^2}}\:dx$ How do I solve this? $$\int _{0+0}^{1-0}\frac{dx}{\left(4-3x\right)\sqrt{x-x^2}}\:dx$$
I know it's a type 3 improper integral, and I'm having issues with these. I think that I need to write it as a sum of limits and then try to compute the values of those limits and the value of the sum would be my improper integral value.
Do I need to try and work it around with a substitution? I was thinking I could use trigonometric substitution for this, but I don't think it applies here.
Can anyone give me a hint or help me with this?
 A: Sub $x=u^2$; then the integral is
$$2 \int_0^1 \frac{du}{(4-3 u^2) \sqrt{1-u^2}} = \int_{-1}^1 \frac{du}{(4-3 u^2) \sqrt{1-u^2}}$$
We can evaluate this using a trig substitution and then another step which I will explain.  Let $u = \sin{t}$; then the integral is
$$\int_{-\pi/2}^{\pi/2} \frac{dt}{4 - 3 \sin^2{t}} $$
which we can rewrite, using the double angle formula and subbing again, as
$$\int_{-\pi}^{\pi} \frac{dt}{5+3 \cos{t}} $$
We may convert this to an integral over the unit circle in the complex plane and use the residue theorem.  Let $z=e^{i t}$ and the integral is
$$-i \oint_{|z|=1} \frac{dz}{z} \frac1{5+\frac32 (z+z^{-1})} = -i 2 \oint_{|z|=1} dz \, \frac1{3 z^2 + 10 z + 3}$$
The poles of the denominator are at $z=-3$ and $z=-1/3$; only $z=-1/3$ is in the unit circle.  The residue theorem states that the integral is equal to $i 2 \pi$ times the residue at that pole, or
$$i 2 \pi (-i 2) \frac1{6(-1/3)+10} = \frac{\pi}{2}$$

Alternatively, if you do not like residues, you could simply use the substitution $v=\tan{(t/2)}$; then the integral becomes
$$\int_{-\infty}^{\infty} \frac{dv}{4+v^2} = \frac{\pi}{2} $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\large A\ \Re\mbox{-}{\it Integration}}$:
\begin{align}
&\bbox[#ffd]{\int_{0}^{1}{\dd x \over \pars{4 - 3x}\root{x - x^{2}}}} =
\int_{\infty}^{1}{-\dd x/x^{2} \over \pars{4 - 3/x}
\root{1/x - 1/x^{2}}}
\\[5mm] = &\
\int_{1}^{\infty}{\dd x \over \pars{4x - 3}\root{x - 1}} =
\int_{0}^{\infty}{\dd x \over \pars{4x + 1}\root{x}}
\\[5mm] = &\
\half\int_{0}^{\infty}{\dd x \over \pars{x + 1}\root{x}}
\end{align}

\begin{align}
&\mbox{With}\quad x = t^{2}:
\\
&\bbox[#ffd]{\int_{0}^{1}{\dd x \over \pars{4 - 3x}\root{x - x^{2}}}} =
\int_{0}^{\infty}{\dd t \over t^{2} + 1} =
\bbox[10px,border:1px groove navy]{\pi \over 2} \\ &
\end{align}
A: \begin{align}
I&=\int_{0}^{1}\!{\frac {1}{ \left( 4-3\,x \right) \sqrt {-{x}^{2}+x}}}
\,{\rm d}x\\
&=\int_{-1/2}^{1/2}\!-4\,{\frac {1}{\sqrt {-4\,{u}^{2}+1} \left( -5+6\,u
 \right) }}\,{\rm d}u,\,\,[\mbox{sub}, u = x-1/2]\\ 
&=-4\,\int_{-1/2}^{1/2}\!{\frac {1}{\sqrt {-4\,{u}^{2}+1} \left( -5+6\,u
 \right) }}\,{\rm d}u\\
&=-4\,\int_{\infty }^{0}\! \left( 8\,{{\it u_1}}^{2}+2 \right) ^{-1}
\,{\rm d}{\it u_1},\,\,\,[\mbox{sub}, u_1 = ((1/2-u)/(1/2+u))^{(1/2)}]\\
&=-4\,\int_{\pi /2}^{0}\!\frac{1}{4}\,{\rm d}{\it u_2},\,\,\,[\mbox{sub}, u_1 = 1/2\tan(u_2)]\\
&=\frac{\pi}{2}
\end{align}
