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