TITLE: Evaluate $\int_0^1 \frac{x^k-1}{\ln x}dx $ using high school techniques
QUESTION [17 upvotes]: Is there a way to compute this integral,  $$\int_0^1 \frac{x^k-1}{\ln x}dx =\ln({k+1})$$ without using the derivation under the integral sign nor transforming it to a double integral and then interchanging the order of integration. High school techniques only if possible.

REPLY [10 votes]: $$
\begin{align}
\int_0^1\frac{x^k-1}{\log(x)}\mathrm{d}x
&=\int_0^\infty\frac{e^{-x}-e^{-(k+1)x}}{x}\mathrm{d}x\tag{1}\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-x}-e^{-(k+1)x}}{x}\mathrm{d}x\tag{2}\\
&=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}{x}\mathrm{d}x-\int_\epsilon^\infty\frac{e^{-(k+1)x}}{x}\mathrm{d}x\right]\tag{3}\\
&=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}{x}\mathrm{d}x-\int_{(k+1)\epsilon}^\infty\frac{e^{-x}}{x}\mathrm{d}x\right]\tag{4}\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^{(k+1)\epsilon}\frac{e^{-x}}{x}\mathrm{d}x\tag{5}\\
&=\lim_{\epsilon\to0^+}\int_1^{k+1}\frac{e^{-\epsilon x}}{x}\mathrm{d}x\tag{6}\\
&=\int_1^{k+1}\frac1x\mathrm{d}x-\lim_{\epsilon\to0^+}\int_1^{k+1}\frac{1-e^{-\epsilon x}}x\mathrm{d}x\tag{7}\\
&=\int_1^{k+1}\frac1x\mathrm{d}x\tag{8}\\[10pt]
&=\log(k+1)\tag{9}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto e^{-x}$
$(2)$: express as a limit so that we can separate the integrals
$(3)$: separate the integrals
$(4)$: substitute $x\mapsto x/(k+1)$
$(5)$: recombine integrals
$(6)$: substitute $x\mapsto\epsilon x$
$(7)$: separate the significant from the rest
$(8)$: Squeeze Theorem: $1-\epsilon x\le e^{-\large\epsilon x}\le1\implies0\le\int_1^{k+1}\frac{1-e^{-\large\epsilon x}}x\mathrm{d}x\le\epsilon k$
$(9)$: integrate