integral problem

I did this one a little differently...there is a "fudge factor" involved, but it's only a slight "fudge"

∫ x / √(3 - 4x^2) dx

let u = 3 - 4x^2    du  = -8x dx    du/-8 = x dx

So we have

-(1/8)∫u^(-1/2) du =

-(1/4)u^(1/2) + C =

-(1/4)(3 - 4x^2)^(1/2) + C

-

.

(2/sqrt3)tan^(-1)(2x/sqrt3)+c

$$\int\;\frac{1}{x^2+\frac{3}{4}}\;dx\;=\;\frac{2}{\sqrt3}\;tan^{-1}\;\frac{2x}{\sqrt3}+c$$

$$\int\;\frac{1}{x^2+a^2}\;dx\;=\;\frac{1}{a}\;tan^{-1}\;\frac{x}{a}+c$$

thank you alan a lot, but i have one more problem with this:

$$$\int \left(\frac{x}{\sqrt{3-4x^2}}\right)dx$$ $

can you help me please??

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.

$$\\v=x\;\;and\;\;u'=\frac{1}{\sqrt{a^2-x^2}}\\\\ v'=1\qquad u=sin^{-1}\;\frac{x}{a}\\\\ \;\int\; x*\frac{1}{\sqrt{a^2-x^2}}\;dx\\\\$$

$$\\=\frac{1}{2}(sin^{-1}\;\frac{x}{a}\;*\;x\;\;-\;\;\int\;sin^{-1}\;\frac{x}{a}\;*\;1\;dx)\\\\ =\frac{1}{2}\left(xsin^{-1}\;\frac{x}{a}\;\;-\left[a\sqrt{1-\frac{x^2}{a^2}}\;+\;xsin^{-1}\;\frac{x}{a}\;\;\right]\right)+c\\\\ =\frac{1}{2}\left(\;-\left[a\sqrt{1-\frac{x^2}{a^2}}\;\right]\right)+c\\\\ =\frac{-1}{2}\left(a\;\sqrt{1-\frac{x^2}{a^2}}\;\right)+c\\\\ =\frac{-1}{2}\left(\frac{\sqrt{3}}{2}\;\sqrt{1-\frac{4x^2}{3}}\;\right)+c\\\\ =\frac{-1}{2}\left(\frac{\sqrt{3}}{2}\;\sqrt{\frac{3-4x^2}{3}}\;\right)+c\\\\ =\frac{-1}{2}\left(\frac{1}{2}\;\sqrt{3-4x^2}\;\right)+c\\\\ =\;\frac{-\sqrt{3-4x^2}}{4}\;+c\\\\$$

$$$\int \left(\frac{x}{\sqrt{3-4x^2}}\right)\ dx \quad \text { ?}$$ $

$$\small{ \text{ $ \begin{array}{rcl} &=&\int \left(\frac{ \big{x} }{\sqrt{3\left(1-\frac{4}{3}x^2\right)}} \right) \ dx \\ \\ &=&\frac{1 }{\sqrt{3}}\int \left(\frac{ \big{x} }{\sqrt{ 1- \left( \frac{ \big{x} }{ \sqrt \frac{3}{4} } \right)^2 } } \right) \ dx \end{array} $ }} $\\\\$ \small\text{ we substitue: $ \frac{x} {\sqrt{ \frac{3}{4} } } = \sin(u) \quad \Rightarrow \quad \frac{ \ dx} {\sqrt{ \frac{3}{4} } } = \cos(u) \ du$ }} $\\\\$ \small\text{ and set also: $ x = ( \sqrt{ \frac{3}{4} } ) * \sin(u) \quad $ and $\quad \ dx = ( \sqrt{ \frac{3}{4} } ) * \cos(u) \ du$ }} $\\\\$ \small\text{ $ =\frac{1 }{\sqrt{3}}\int \left(\frac{ ( \big{ \sqrt{ \frac{3}{4} } ) * \sin(u) } }{\sqrt{ 1- \big{ \left( \sin(u) \right)^2 } } } \right) ( \sqrt{ \frac{3}{4} } ) * \cos(u) \ du$ }}$$

$$$\\\\$ \small\text{ $ =\frac{1 }{\sqrt{3}}\int \left( \frac{ ( \big{ \sqrt{ \frac{3}{4} } ) * \sin(u) } } { \big{\cos(u) } } \right) ( \sqrt{ \frac{3}{4} } ) * \cos(u) \ du$ }} $\\\\$ \small\text{ $ =\dfrac{\frac{3}{4} }{ \sqrt{3} }\int \left( \sin(u) \ du$ }} $\\\\$ \small\text{ $ =\frac{ \sqrt{3}}{4}\int \left( \sin(u) \ du$ }} $\\\\$ \small\text{ $ =\frac{ \sqrt{3}}{4}\int \left( \sin(u) \ du \quad | \quad \int\sin(u)\ du = -\cos(u)$ }} $\\\\$ \small\text{ $ =\frac{ \sqrt{3}}{4}(-\cos(u))$ }} $\\\\$ \small\text{ $ =-\frac{ \sqrt{3}}{4}\cos(u) \quad | \quad cos(u) = \sqrt{1-\sin(u)^2 }= \sqrt{1-\frac{4}{3}x^2 } $ }} $\\\\$ \small\text{ $ =-\frac{ \sqrt{3}}{4}\sqrt{1-\frac{4}{3}x^2 } }$ }} $\\\\$ \small\text{ $ =-\frac{1}{4}\sqrt{3-4x^2 } }$ }}$$

$$\boxed{\int \left(\frac{x}{\sqrt{3-4x^2}}\right)\ dx =-\frac{1}{4}\sqrt{3-4x^2 } \quad + c }$$

Chris, I really like what you have done.  It is really neat and looks perfectly valid too me.  

Heureka's answer is also much better than mine.   Thanks Heureka