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I have a question where we need to find an integral using where "$u = 1+e^{x}$" for the equation "$\int \frac{e^{3x}}{1+e^{x}}dx$".

However when I substitute it I end up with "$\int \frac{(u-1)^{3}}{u}du$" instead of "$\int \frac{(u-1)^{2}}{u}du$" which is what I should be getting. Please help

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  • $\begingroup$ You are correct, if $u=1+e^x$, then $e^x=u-1$ and $e^{3x}=(u-1)^3$. Is it possible whatever solutions you are looking at are incorrect? $\endgroup$ – kccu Jan 21 at 0:22
  • $\begingroup$ you should include $dx$ and $du$ $\endgroup$ – J. W. Tanner Jan 21 at 0:24
  • $\begingroup$ Have checked on two websites and both somehow have $e^{3x}=(u−1)^{2}$ and they get the same answer as the one on in the answers section. $\endgroup$ – P.Lord Jan 21 at 0:24
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$$u=1+e^x \to du= e^x dx \to dx =\frac{du}{u-1}$$ Also $e^{3x}=(u-1)^3$

Put it together and we have:

$$\int \frac{e^{3x}}{1+e^x}\ dx = \int \frac{(u-1)^3}{u}\cdot \frac{du}{u-1} =\int \frac{(u-1)^2}{u} \ du $$ as required

Your mistake was that you didnt substitute in $du$ in for $dx$ when you applied the u-sub.

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  • $\begingroup$ Thank you so much you are a life saver. $\endgroup$ – P.Lord Jan 21 at 0:28
  • $\begingroup$ You're most welcome. Thank you for the good question. $\endgroup$ – Rhys Hughes Jan 21 at 0:31

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