Integration problem

∈=p/Ao((1-(x/2L) *E^(-1)

I am trying to integrate the above. For clarity p is over Ao((1-(x/2L) and then all multiplied by E^(-1). Do I need to deal with the Ao((1-(x/2L) first?

Thanks in advance Paul

Hi Paul, I understand you need to integrate the equation but you did not present to us the variable of integration. Assuming your variable is x then other symbol like p, Ao, L, and E are constants. Am I right? If so, is this your equation?
$\displaystyle \epsilon = \int \dfrac{pE^{-1}}{A_o \left( \dfrac{1 - x}{2L} \right)} \, dx$

Assuming my interpretations are correct, here is how to integrate it:
$\displaystyle \epsilon = \int \dfrac{pE^{-1}}{A_o \left( \dfrac{1 - x}{2L} \right)} \, dx$

$\displaystyle \epsilon = \dfrac{2pL}{EA_o}\int \dfrac{dx}{1 - x}$

$\displaystyle \epsilon = -\dfrac{2pL}{EA_o}\ln (1 - x) + C$

In reply to by Jhun Vert

Hi Romel, you are correct variable is x. I couldn't paste the equation in. The e-1 should be multiplied by the p/A0 equation, but I think I understand it. I need to run some software now and compare the hand calcs.

The other way that I looked at your equation is this:
$\displaystyle \epsilon = \int \dfrac{pE^{-1}}{A_o \left(1 - \dfrac{x}{2L} \right)} \, dx$

$\displaystyle \epsilon = \dfrac{p}{EA_o}\int \dfrac{dx}{\dfrac{2L - x}{2L}}$

$\displaystyle \epsilon = \dfrac{2pL}{EA_o}\int \dfrac{dx}{2L - x}$

$\displaystyle \epsilon = -\dfrac{2pL}{EA_o}\ln (2L - x) + C$

If you copy and paste the flicker link you will see the equation I am trying to solve Romel