# Stuck with Integration by Substitution

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

• 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? – kccu Jan 21 at 0:22
• you should include $dx$ and $du$ – J. W. Tanner Jan 21 at 0:24
• 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. – P.Lord Jan 21 at 0:24

$$u=1+e^x \to du= e^x dx \to dx =\frac{du}{u-1}$$ Also $$e^{3x}=(u-1)^3$$
$$\int \frac{e^{3x}}{1+e^x}\ dx = \int \frac{(u-1)^3}{u}\cdot \frac{du}{u-1} =\int \frac{(u-1)^2}{u} \ du$$ as required
Your mistake was that you didnt substitute in $$du$$ in for $$dx$$ when you applied the u-sub.