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December 28th, 2014, 06:32 AM   #1
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Simplify

How to simplify this expression?
$\displaystyle \frac{1}{x}\cdot \sqrt{4+3x}+\frac{3}{2\sqrt{3x+4}}\cdot\ln x$

I know the answer is $\displaystyle \frac{3x(\ln(x)+2)+8}{2x\sqrt{3x+4}}$. But how?

Last edited by skipjack; December 28th, 2014 at 07:48 PM.
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December 28th, 2014, 07:15 AM   #2
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The common denominator is $\displaystyle \ 2x\sqrt{3x+4}$

The first fraction is amplified by $\displaystyle \ 2\sqrt{3x+4}$

and the second fraction is amplified by $\displaystyle \ x$

Thus, we get

$\displaystyle \color{blue}{\dfrac{2\sqrt{3x+4}\cdot\sqrt{3x+4}+3 x \ln x}{2x \sqrt{3x+4}}=\dfrac{2(3x+4)+3x \ln x}{2x \sqrt{3x+4}}=\dfrac{2\cdot3x+8+3x \ln x}{2x \sqrt{3x+4}}
\\\;\\
=\dfrac{3x(2+ \ln x) + 8}{2x \sqrt{3x+4}}}$

Last edited by skipjack; December 28th, 2014 at 07:51 PM.
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