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

  • $\begingroup$ 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

$$u=1+e^x \to du= e^x dx \to dx =\frac{du}{u-1}$$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.