We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+5
863
7
avatar+16 

$$$\int \frac{1}{\left(x^2+\frac{3}{4}\right)}dx$$$

 

Somobody who know how to solve this?? Please :)

 Dec 7, 2014

Best Answer 

 #6
avatar+104855 
+10

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

 

 

 

-

 Dec 8, 2014
 #1
avatar+28188 
+10

Integral:

 

.

 Dec 7, 2014
 #2
avatar+105634 
+5

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

 

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

 

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

.
 Dec 7, 2014
 #3
avatar+16 
0

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??

 Dec 7, 2014
 #4
avatar+105634 
+10

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.

 

 

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

 

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

.
 Dec 8, 2014
 #5
avatar+23317 
+10

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

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

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

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

.
 Dec 8, 2014
 #6
avatar+104855 
+10
Best Answer

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

 

 

 

-

CPhill Dec 8, 2014
 #7
avatar+105634 
0

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    

 Dec 9, 2014

33 Online Users

avatar
avatar
avatar
avatar