When letting $u=e^y$, why is it possible to substitute $e^{2y}$ with $u$ as well?

Question

I'm not sure if I'm forgetting some properties regarding exponents, but anyway. I went to WolframAlpha to verify my answer on the indefinite integral $\int{\frac{e^{2y}}{e^y-1}}\,dy$. Though I'm missing something since Wolfram used $u$ substitution with $u=e^y$. Now when substituting $u$ into the equation it wrote $\int{\frac{u}{u-1}}\,du$. This got me really confused, since, to me this means that $e^y=e^{2y}$.

Answer 1

$$\int\frac{e^{2y}}{e^y-1}\,dy=\int\frac{e^y}{e^y-1}\color{red}{(e^y\,dy)}=\int\frac u{u-1}\,du$$

Answer 2

If you do $u=e^y$, then you must also do $\mathrm du=e^y\,\mathrm dy$. So,$$\int\frac{e^{2y}}{e^y-1}\,\mathrm dy\left(=\int\frac{(e^y)^2}{e^y-1}\,\mathrm dy=\int\frac{e^y}{e^y-1}e^y\,\mathrm dy\right)$$becomes $$\int\frac u{u-1}\,\mathrm du.$$

Answer 3

Let $u=e^y$. Then $\frac{du}{dy}=\frac{d}{dy}e^y=e^y$, so $dy=\frac{du}{e^y}$. Now exchange $dy$ for $\frac{du}{e^y}$ and notice that

$$\int \frac{e^{2y}} {e^y-1} dy = \int \frac{e^{2y}} {e^y-1} \frac{du}{e^y} = \int \frac{e^{2y}} {e^y} \cdot \frac{1}{e^y-1} du = \int \frac{e^y} {1} \cdot \frac{1}{e^y-1} du = \int \frac{e^y} {e^y-1} du$$

Now exchange $u$ for $e^y$ as follows

$$\int \frac{u} {u-1} du$$