Correct answer, wrong method in improper integral problem. Note: this is a problem from a class, but my solution has already been submitted and the assignment is locked for all students so this shouldn't break any sort of academic honesty guidelines.
$$\int_{0}^{\ln{3}}\frac{e^x}{(e^x-1)^\frac{2}{3}}dx$$
I understand that this is an improper integral given $\frac{e^0}{(e^0 - 1)^\frac{2}{3}} = \frac{1}{0}$, but I seem to have come to the right answer without using any improper integration techniques. Through two layers of u-substitution $u = e^x$ and $v = u - 1$, my answer comes to $3 \cdot 2^\frac{1}{3}$. Every online calculator I've checked with confirms that this is correct, but I don't understand why it seems to be correct without any use of improper integration techniques. This hints at some disconnect I've got with improper integrals or u-substitution, and I'm really at a loss for what it might be.
Thanks for any insight!
 A: Since the integral is improper at the lower bound, it is defined as $\lim_{\epsilon\to0} \int_\epsilon^{\ln 3}$. Let us look at what happens:
$$
\int_{\epsilon}^{\ln{3}}\frac{e^x}{(e^x-1)^\frac{2}{3}}dx
= \{ u=e^x \}
= \int_{e^\epsilon}^{3} \frac{u}{(u-1)^\frac{2}{3}} \frac{du}{u}
= \int_{e^\epsilon}^{3} \frac{du}{(u-1)^\frac{2}{3}} \\
= \{v=u-1\} 
= \int_{e^\epsilon-1}^{2} \frac{dv}{v^\frac{2}{3}}
= \int_{e^\epsilon-1}^{2} v^{-\frac{2}{3}} \, dv
= \left[ \frac{1}{3}v^{\frac{1}{3}} + C \right]_{e^\epsilon-1}^{2} \\
= \frac{1}{3} \cdot 2^{\frac{1}{3}} - \frac{1}{3} (e^\epsilon-1)^{\frac{1}{3}}
.
$$
Taking limits we find that
$\frac{1}{3} (e^\epsilon-1)^{\frac{1}{3}} \to 0 = \frac{1}{3} (e^0-1)^{\frac{1}{3}}$
as $\epsilon \to 0,$ i.e. the primitive function is defined and continuous at $0$. And that is the reason it works.
Another example is the improper integral $\int_0^1 \ln x \, dx$. For this we get
$$
\int_\epsilon^1 \ln x \, dx = \left[ x\ln x - x \right]_\epsilon^1
= (1\ln 1-1) - (\epsilon\ln\epsilon-\epsilon).
$$
Here the lower bound term is not defined for $\epsilon=0$ and we can not just insert $\epsilon=0$ but really need to take limits (which result in $0$ so the full integral has value $-1$).
A: We can perform a somewhat different substitution of the form $$u^{-3} = e^x - 1, \quad -3u^{-4} du = e^x \, dx.$$  This maps the endpoints of integration from $x = 0$ to $u = \infty$ and $x = \log 3$ to $u = 2^{-1/3}$, and the integral becomes
$$\int_{u=\infty}^{2^{-1/3}} u^2 (-3u^{-4}) \, du = 3 \int_{u=2^{-1/3}}^\infty u^{-2} \, du.$$  Then we can apply the usual procedures:  $$3 \int_{u=2^{-1/3}}^\infty u^{-2} \, du = 3 \lim_{N \to \infty} \int_{u=2^{-1/3}}^N u^{-2} \, du = 3\lim_{N \to \infty} \left[- \frac{1}{u} \right]_{u=2^{-1/3}}^N = 3(2^{1/3}) - 3\lim_{N \to \infty} \frac{1}{N} = 3(2^{1/3}).$$
A: Let $u = e^x -1$.  Then $du = e^x\, dx$.  Your integral becomes
$$\int_0^2 {du\over u^{2/3}}.$$
This integral integrates at zero because the power is less than 1.
