Evaluate the indefinite integral with u substitution

Question

$ \displaystyle\frac{ e^{2x}}{e^x+1} $ So I'm currently stuck on this problem. I looked at my teachers solution and they solved it using u substitution by setting $u = e^x +1$
I understand this step except I don't get how this helps us at all since $du = e^xdx$ unless I have something incorrect. With that in mind how does this address $ e^{2x} $. How do I solve this and get rid of the $e^{2x} $ with u substitution?

Answer 1

You want $$ \int \frac{\mathrm{e}^{2x}}{\mathrm{e}^x + 1} \,\mathrm{d}x \text{.} $$ The suggestion is to use $u = \mathrm{e}^x+1$ so that $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x$. That gives us the following gibberish (never write this variety of Frankenstein's monster where anyone else will ever see it): $$ \int \frac{\mathrm{e}^{x}}{u} \,\mathrm{d}u \text{.} $$ Notice, though, that the part of the integral we haven't converted to be in terms of $u$, $\mathrm{e}^x$, is $u - 1$, so we can finish the substitution: $$ \int \frac{u-1}{u} \,\mathrm{d}u \text{.} $$