Some careless errors have been fixed - Thanks very much Alan.
It is correct now.
I have been perfecting my trademark here.
If there is a R E A L L Y L O N G W A Y to do something I WILL FIND IT.
In my opinion CPhill's answer is the best one here.
It is really elegant. Thanks Chris.
Heureka's answer is also much better than mine. Thanks Heureka 
$$\int \left(\frac{x}{\sqrt{3-4x^2}}\right)dx\\\\
=\int \left(\frac{x}{\sqrt{4*(\frac{3}{4}-x^2)}}\right)dx\\\\
=\int \frac{x}{2\sqrt{(\frac{3}{4}-x^2)}}\;dx\\\\
=\frac{1}{2}\;\int \frac{x}{\sqrt{(\frac{3}{4}-x^2)}}\;dx\\\\
=\frac{1}{2}\;\int \frac{x}{\sqrt{(\frac{\sqrt{3}}{2})^2-x^2}}\;dx\\\\$$
$$\\=\frac{1}{2}\;\int \frac{x}{\sqrt{a^2-x^2}}\;dx\qquad where \quad a=\frac{\sqrt3}{2}\\\\
=\frac{1}{2}\;\int\;vu' \;dx\qquad where \quad v=x\;\;and\;\;u'=\frac{1}{\sqrt{a^2-x^2}}\\\\$$
Now use integration by parts to solve.
$$\\v=x\;\;and\;\;u'=\frac{1}{\sqrt{a^2-x^2}}\\\\
v'=1\qquad u=sin^{-1}\;\frac{x}{a}\\\\
\;\int\; x*\frac{1}{\sqrt{a^2-x^2}}\;dx\\\\$$
$$\\=\frac{1}{2}(sin^{-1}\;\frac{x}{a}\;*\;x\;\;-\;\;\int\;sin^{-1}\;\frac{x}{a}\;*\;1\;dx)\\\\
=\frac{1}{2}\left(xsin^{-1}\;\frac{x}{a}\;\;-\left[a\sqrt{1-\frac{x^2}{a^2}}\;+\;xsin^{-1}\;\frac{x}{a}\;\;\right]\right)+c\\\\
=\frac{1}{2}\left(\;-\left[a\sqrt{1-\frac{x^2}{a^2}}\;\right]\right)+c\\\\
=\frac{-1}{2}\left(a\;\sqrt{1-\frac{x^2}{a^2}}\;\right)+c\\\\
=\frac{-1}{2}\left(\frac{\sqrt{3}}{2}\;\sqrt{1-\frac{4x^2}{3}}\;\right)+c\\\\
=\frac{-1}{2}\left(\frac{\sqrt{3}}{2}\;\sqrt{\frac{3-4x^2}{3}}\;\right)+c\\\\
=\frac{-1}{2}\left(\frac{1}{2}\;\sqrt{3-4x^2}\;\right)+c\\\\
=\;\frac{-\sqrt{3-4x^2}}{4}\;+c\\\\$$
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+118733 
