Evaluate $\int_0^1 \frac{x^k-1}{\ln x}dx $ using high school techniques 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.
 A: Note that $x^k = e^{k \ln(x)}$. Hence, we have
$$\dfrac{x^k-1}{\ln(x)} = \sum_{l=1}^{\infty} \dfrac{k^l \ln^{l-1}(x)}{l!}$$
Hence, our integral becomes
$$\sum_{l=1}^{\infty} \dfrac{k^l}{l!} \int_0^1 \ln^{l-1}(x)dx = \sum_{l=1}^{\infty} \dfrac{k^l}{l!} \int_0^1 \ln^{l-1}(x)dx = \sum_{l=1}^{\infty} (-1)^{l-1}\dfrac{k^l}l = \ln(1+k)$$
since $$\displaystyle \int_0^1 \ln^{l-1}(x)dx = \int_{\infty}^0 (-t)^{l-1} e^{-t} (-dt)= (-1)^{l-1}\int_0^{\infty} t^{l-1}e^{-t} dt = (-1)^{l-1} (l-1)!$$

As pointed out in the comments, this is true for $-1<k\leq 1$. For $k> 1$, the proof can be easily extended. Set $$I(k) = \int_0^1 \dfrac{x^k-1}{\ln(x)}dx$$
We then have
\begin{align}
I(k) - I(k-1) & = \int_0^1 x^{k-1}\dfrac{(x-1)}{\ln(x)}dx = \sum_{l=1}^{\infty}\int_0^1 \dfrac{x^{k-1} \ln^{l-1}(x)}{l!}dx\\
& = \sum_{l=1}^{\infty}(-1)^l \dfrac{(1/k)^l}l = \ln(1+1/k) \text{ (since $1/k<1$)}
\end{align}
This now gives us $I(k) = \ln(1+k)$ for all $k$.
A: 
I'm not expecting any upvote for this but surely this answer could also help someone!.

Let
$$I(k)=\int_0^1{\frac{x^k-1}{\ln x}dx}$$
$$I'(k)=\int_0^1{\frac{(x)^k\ln x}{\ln x}dx}=\int_0^1{x^kdx}=\frac{x^{k+1}}{k+1}=\frac{1}{k+1}$$
$$I(k)=\ln(k+1)+C$$
Since we have $I(0)=0\implies \ln(1)+C=0\implies C=0$

$$I(k)=\int_0^1{\frac{x^k-1}{\ln x}dx}=\ln(k+1)$$

A: Yet another answer.
It is easy to see that $\displaystyle f(x)=\frac{x^k-1}{\log x}$, with $f(0)=0$ and $f(1)=k$ is a continuous fonction on $[0,1]$ for $k> 0$. Put now for $0<h<1$ $\displaystyle F(h)=\int_0^h f(t)dt$. We have $F(h)\to F(1)=I_k$ if $h\to 0$, $h<1$.
But:
$$F(h)=\int_0^{h}\frac{x^k}{\log x}dx-\int_0^h \frac{dx}{\log x}=\int_{h^{k+1}}^h \frac{-dx}{\log x}$$
By the change of variable $t^{k+1}=u$ in the first integral.
Now for $x\in [h^{k+1},h]$ we have
$$h^{k+1}\frac{-1}{x\log x}\leq \frac{-x}{x\log x}\leq h\frac{-1}{x\log x}$$
where we have used that $\displaystyle \frac{-1}{x\log x}\geq 0$.
As $h^{k+1}\leq h$, integrating we get:
$$h^{k+1}\int_{h^{k+1}}^h \frac{-1}{x\log x}dx\leq F(h)\leq h\int_{h^{k+1}}^h\frac{-1}{x\log x}dx$$
Hence
$$h^{k+1}\log(k+1)\leq F(h)\leq h\log(k+1)$$
and if $h\to 1$, we get the result. 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{1}{x^{k} - 1 \over \ln\pars{x}}\,\dd x =\ln\pars{k + 1}:\
     {\large ?}}$.

\begin{align}&\color{#66f}{\large\int_{0}^{1}{x^{k} - 1 \over \ln\pars{x}}\,\dd x}
=-\int_{0}^{1}\pars{x^{k} - 1}\
\overbrace{\int_{0}^{\infty}x^{t}\,\dd t}^{\ds{=\ \color{#c00000}{-\,{1 \over \ln\pars{x}}}}}\ \,\dd x\ =\
-\int_{0}^{\infty}\int_{0}^{1}\pars{x^{k + t} - x^{t}}\,\dd x\,\dd t
\\[5mm]&=\int_{0}^{\infty}\pars{{1 \over t + 1} - {1 \over t + k + 1}}\,\dd t
=\left.\ln\pars{t + 1 \over t + k + 1}\right\vert_{\, t\ =\ 0}^{\, t\ \to\ \infty}
=-\ln\pars{1 \over k + 1}
\\[5mm]&=\color{#66f}{\large\ln\pars{k + 1}}
\end{align}
A: Here is a seemingly elementary solution: Using the following simple limit
$$ \ln x = \lim_{h\to 0} \frac{x^{h} - 1}{h} = \lim_{n\to\infty} \frac{x^{1/n} - 1}{1/n}, $$
it follows that
\begin{align*}
\int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx
&= \lim_{n\to\infty} \int_{0}^{1} \frac{x^{k} - 1}{n (x^{1/n} - 1)} \, dx \\
&= \lim_{n\to\infty} \int_{0}^{1} \frac{(y^{nk} - 1)y^{n-1}}{y - 1} \, dy, \quad (\text{substitute } x = y^{n}) \\
&= \lim_{n\to\infty} \int_{0}^{1} (1+y+\cdots +y^{nk-1})y^{n-1} \, dy.
\end{align*}
Now using the following simple formula
$$ \int_{0}^{1} y^{i-1} \, dy = \frac{1}{i}, $$
we get the following identity
\begin{align*}
\int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx
&= \lim_{n\to\infty} \left( \frac{1}{n} + \cdots + \frac{1}{nk+n-1} \right) 
 =\lim_{n\to\infty} \sum_{i=0}^{nk-1} \frac{1}{n+i}.
\end{align*}
Modifying the summand a little bit, we have
\begin{align*}
\int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx
&= \lim_{n\to\infty} \sum_{i=0}^{nk-1} \frac{1}{n} \frac{1}{1+(i/n)}
 = \int_{0}^{k} \frac{dt}{1+t}
 = \ln(k+1).
\end{align*}
P.s. All these steps are elementary except for one point. Can you figure out where it is?

EDIT. Elementary solution is not always the easiest solution as the essence is often shadowed by the technical intricacy involved. Nevertheless, here is a solution which uses only freshman-level calculus:
I will assume that $k$ is a positive integer (as I implicitly did in the first solution). Now let $n$ be another positive integer. Then by the substitution $x = y^{n}$ we get
$$ I := \int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx = \int_{0}^{1} \frac{(y^{nk} - 1)y^{n-1}}{\log y} \, dy. $$
For the solution, we claim the following:

Claim. For $0 < y < 1$ we have $y \leq \frac{y-1}{\log y} \leq 1$.
Proof. Let $f(y) = y \log y + 1 - y$ and $g(y) = y - 1 - \log y$. By differentiating once, we have
  $$ f'(y) = \log y \leq 0 \quad \text{and} \quad g'(y) = 1 - y^{-1} \leq 0.$$
  This shows that both $f$ and $g$ are decreasing on $(0, 1]$. Since $f(1) = 0$ and $g(1) = 0$, we get both
  $$ f(y) \geq 0 \quad \text{and} \quad g(y) \geq 0 \quad \text{for} \quad 0 < y < 1. $$
  The claim follows from these two inequalities. ////

Now write $I$ in the following form:
$$ I = \int_{0}^{1} \frac{(1 - y^{nk})y^{n-1}}{1 - y} \frac{y - 1}{\log y} \, dy. $$
Then by the claim we obtain
$$ \int_{0}^{1} \frac{(1 - y^{nk})y^{n}}{1 - y} \, dy \leq I \leq \int_{0}^{1} \frac{(1 - y^{nk})y^{n-1}}{1 - y} \, dy. $$
We know how to calculate both integrals and where they converge as $n \to \infty$ from the previous answer:
$$ \lim_{n\to\infty} \int_{0}^{1} \frac{(1 - y^{nk})y^{n}}{1 - y} \, dy
= \lim_{n\to\infty} \int_{0}^{1} \frac{(1 - y^{nk})y^{n-1}}{1 - y} \, dy
= \log(k+1). $$
Therefore the conclusion follows from the Squeezing Theorem.
A: $$
\begin{align}
\int_0^1\frac{x^k-1}{\log(x)}\mathrm{d}x
&=\int_0^\infty\frac{e^{-x}-e^{-(k+1)x}}{x}\mathrm{d}x\tag{1}\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-x}-e^{-(k+1)x}}{x}\mathrm{d}x\tag{2}\\
&=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}{x}\mathrm{d}x-\int_\epsilon^\infty\frac{e^{-(k+1)x}}{x}\mathrm{d}x\right]\tag{3}\\
&=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}{x}\mathrm{d}x-\int_{(k+1)\epsilon}^\infty\frac{e^{-x}}{x}\mathrm{d}x\right]\tag{4}\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^{(k+1)\epsilon}\frac{e^{-x}}{x}\mathrm{d}x\tag{5}\\
&=\lim_{\epsilon\to0^+}\int_1^{k+1}\frac{e^{-\epsilon x}}{x}\mathrm{d}x\tag{6}\\
&=\int_1^{k+1}\frac1x\mathrm{d}x-\lim_{\epsilon\to0^+}\int_1^{k+1}\frac{1-e^{-\epsilon x}}x\mathrm{d}x\tag{7}\\
&=\int_1^{k+1}\frac1x\mathrm{d}x\tag{8}\\[10pt]
&=\log(k+1)\tag{9}
\end{align}
$$
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
