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paulmdrdo
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- May 13, 2013
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how would start solving this
$\displaystyle\int\frac{dy}{1+e^y}$
$\displaystyle\int\frac{dy}{1+e^y}$
[tex]\displaystyle \int\frac{dy}{1+e^y}[/tex]
Alternatively, MULTIPLY top and bottom by [tex]\displaystyle \begin{align*} e^y \end{align*}[/tex], givinghow would start solving this
$\displaystyle\int\frac{dy}{1+e^y}$
Am I correct in assuming that you meant [tex]\displaystyle \begin{align*} \int{ \frac{e^{2x}}{1 + e^x}\,dx} \end{align*}[/tex]?how did you know that you have to divide the numerator and denominator by e^y?
what if we have
$\displaystyle\int\frac{e^2x}{1+e^x}dx$
yes.sorry for typo error!Alternatively, MULTIPLY top and bottom by [tex]\displaystyle \begin{align*} e^y \end{align*}[/tex], giving
[tex]\displaystyle \begin{align*} \int{ \frac{e^y\,dy}{e^y + \left( e^y \right) ^2}} \end{align*}[/tex]
which gives a form that a very simple substitution of [tex]\displaystyle \begin{align*} u = e^y \implies du = e^y\,dy \end{align*}[/tex] should be able to handle.
How did we know to do this? Well, since there's an [tex]\displaystyle \begin{align*}e^y \end{align*}[/tex] function, and we know that it's its own derivative, that means if we were to make any substitution using it, it also needs to appear as a factor, as when we do an integration by substitution, the derivative of the function we are substituting needs to be a factor of the entire function being integrated.
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Am I correct in assuming that you meant [tex]\displaystyle \begin{align*} \int{ \frac{e^{2x}}{1 + e^x}\,dx} \end{align*}[/tex]?
Here you don't need to do any manipulation to the integral at all, notice thatyes.sorry for typo error!
mark my answer is positive $\displaystyle ln\left(\frac{e^y}{e^y+1}\right)+C$ not $\displaystyle -ln\left(\frac{e^y}{e^y+1}\right)+C$ why is that?Using the properties of logs, we may transform the form given by soroban into that which you found:
\(\displaystyle -\ln\left(e^{-y}+1 \right)+C=-\ln\left(\frac{e^y+1}{e^y} \right)+C=-\ln\left(\frac{e^y}{e^y+1} \right)+C\)
I have edited my post above...I should have removed the negative sign when I inverted the argument of the log function...mark my answer is positive $\displaystyle ln\left(\frac{e^y}{e^y+1}\right)+C$ not $\displaystyle -ln\left(\frac{e^y}{e^y+1}\right)+C$ why is that?