Hints on calculating the integral $\int_0^1\frac{x^{19}-1}{\ln x}\,dx$ I would be happy to get some hints on the following integral:
$$
\int_0^1\frac{x^{19}-1}{\ln x}\,dx
$$
 A: Differentiation of the integrand $$f(x,a) = \frac{x^a-1}{\log x}$$ with respect to $a$ gives $$\frac{\partial f}{\partial a} = x^a.$$  Therefore, $$I(a) = \int_{x=0}^1 f(x,a) \, dx$$ implies $$\frac{d I}{d a} = \int_{x=0}^1 x^a \, dx = \frac{1}{a+1}, \quad a > -1.$$  Integrating with respect to $a$ then yields $$I(a) = \log(a+1), \quad a > -1.$$  There are some omitted details, but this is an outline of the general solution.
A: Here is the details from @heropup's answer.
Let us generalize the problem. We will evaluate
$$
\mathcal{I}(\alpha)=\int_0^1\frac{x^\alpha-1}{\ln x}\ dx\qquad;\qquad \alpha>-1.\tag1
$$
Now we apply Feynman's method (differentiate under the integral sign). Diferentiating both sides of $(1)$ yields
\begin{align}
\frac{\partial\mathcal{I}}{\partial\alpha}&=\int_0^1\frac{\partial}{\partial\alpha}\left[\frac{x^\alpha-1}{\ln x}\right]\ dx\\
\mathcal{I}'(\alpha)&=\int_0^1 x^\alpha\ dx\\
&=\left.\frac{x^{\alpha+1}}{\alpha+1}\right|_{x=0}^1\\
&=\frac{1}{\alpha+1}.\tag2
\end{align}
Integrating $(2)$ yields
\begin{align}
\mathcal{I}(\alpha)&=\int\frac{1}{\alpha+1}\ d\alpha\\
&=\ln(\alpha+1)+C.\tag3
\end{align}
In order to find out our constant of integration, we let $\alpha = 0$ so that our integrand is $0$, implying that $C = 0$. Letting $\alpha = 19$ will of course solve our original problem:
\begin{align}
\color{purple}{\int_0^1\frac{x^\alpha-1}{\ln x}\ dx}&\color{purple}{=\ln(\alpha+1)}\\
\int_0^1\frac{x^{19}-1}{\ln x}\ dx&=\large\color{blue}{\ln20}.
\end{align}
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\color{#66f}{\large\int_{0}^{1}{x^{19} - 1 \over \ln\pars{x}}\,\dd x}&=
\int_{0}^{1}\pars{x^{19} - 1}\ \overbrace{\pars{-\int_{0}^{\infty}x^{t}\,\dd t}}
^{\ds{=\ {1 \over \ln\pars{x}}}}\ \,\dd x
=\int_{0}^{\infty}\int_{0}^{1}\pars{x^{t} - x^{t + 19}}\,\dd x\,\dd t
\\[5mm]&=\int_{0}^{\infty}\pars{{1 \over t + 1} - {1 \over t + 20}}\,\dd t
=\left.\ln\pars{t + 1 \over t + 20}\right\vert_{\,t\ =\ 0}^{\,t\ \to\ \infty}
=0 - \ln\pars{1 \over 20}
\\[8mm]&
=\color{#66f}{\Large\ln\pars{20}} \approx {\tt 2.9957}
\end{align}

In addition, you can see
  this method.

A: You may like this method. Note
$$ \lim_{n\to\infty} n(x^{\frac{1}{n}}-1)=\ln x, \text{ for }x>0 $$
and hence, for $m\in\mathbb{N}$,
\begin{eqnarray}
\int_0^1\frac{x^m-1}{\ln x}dx&=&\int_0^1\sum_{i=0}^{m-1}\frac{x^i(x-1)}{\ln x}dx\\
&=&\int_0^1\sum_{i=0}^{m-1}\lim_{n\to\infty}\frac{x^i(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\
&=&\lim_{n\to\infty}\sum_{i=0}^{m-1}\int_0^1\frac{x^i(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\
&=&\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{m-1}\int_0^1x^i\sum_{k=0}^{n-1}x^{\frac{k}{n}}dx\\
&=&\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{m-1}\int_0^1\sum_{k=0}^{n-1}x^{\frac{k}{n}+i}dx\\
&=&\lim_{n\to\infty}\sum_{i=0}^{m-1}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{k}{n}+i+1}\\
&=&\sum_{i=0}^{m-1}\int_0^1\frac{1}{x+i+1}dx\\
&=&\sum_{i=0}^{m-1}\ln\frac{i+2}{i+1}\\
&=&\ln(m+1).
\end{eqnarray}
So for $m=19$, we have
$$ \int_0^1\frac{x^{19}-1}{\ln x}dx=\ln20.$$
A: Let $x = e^{-y}$, we have
$$\int_0^1 \frac{x^{19} - 1}{\log x} dx = \int_0^\infty \frac{e^{-\color{blue}{1}y} - e^{-\color{orange}{20}y}}{y} dy$$
This is in the form of a Frullani's integral and one can read off the value of the integral as
$$( \color{red}{1} - \color{green}{0} )\log\left(\frac{\color{orange}{20}}{\color{blue}{1}}\right) = \log 20
\quad\text{ since }\quad e^{-y} = \begin{cases}\color{red}{1}, &y = 0\\ \color{green}{0}, & y \to \infty\end{cases}$$
If you really need to perform the integral yourself without using Frullani's integral directly, I'll recommend you look at answers of this question and learn the various proof there. 
A good exercise is translate the proof there to your particular case. This will get you familiar with the steps that need to evaluate this sort of integral.
A: Hint: Substitute $\ln x=t$. Use integration by parts formula.
