Integration issue

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vipa2000
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Integration issue

After deriving an equation I have ended up with equation where I have a complex denominator and I am not sure how to simplify. I am trying to work out as a function of x. Equation link is below. Thanks in advance.

https://www.flickr.com/photos/baldypaul/32722700004/in/datetaken/

Jhun Vert
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$\begin{align}
\dfrac{p}{\dfrac{Ao\left( \dfrac{1 - x}{2L} \right)}{E}} & = \dfrac{2pEL}{Ao(1 - x)} \\
& = \dfrac{2pEL}{Ao} \cdot \dfrac{1}{1 - x}
\end{align}$
 

You posted this with a title "integration issue" under calculus, which I presume you wanted to do this:
$\displaystyle \dfrac{2pEL}{Ao} \int \dfrac{dx}{1 - x} = -\dfrac{2pEL}{Ao}\ln (1 - x) + C$
 

Side Note:

Allow me to bring up your previous post which is a little similar to the equation in this post.
 
integral_016-integral-given.gif

 

The only difference is E.
 

Note that
$\dfrac{p}{\dfrac{Ao\left( \dfrac{1 - x}{2L} \right)}{E}} = \dfrac{p}{Ao\left( \dfrac{1 - x}{2L} \right)} * E$

while
$\dfrac{p}{Ao\left( \dfrac{1 - x}{2L} \right)} * E^{-1} = \dfrac{p}{AoE\left( \dfrac{1 - x}{2L} \right)}$
 

I am not sure though if this side note is still relevant to you.

vipa2000
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Romel, this is excellent.

If you get chance, could you explain some of the steps i.e. not sure how E is positive and a numerator, and in addition the ln(1-x). Thanks in advance Paul