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Question

Encontrar $\int_0^1 \mathrm{\frac{x-1}{ln(x)}}\,\mathrm{d}x$

He intentado esto:

$\int_0^1 \mathrm{\frac{x-1}{ln(x)}} = \int_0^1 \mathrm{\frac{x}{ln(x)}} - \int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x$

A $\int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x$

No $t=lnx $ $\frac{dt}{dx}=\frac{1}{x}$ y $dx=e^tdt$

$\int \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x \equiv \int \mathrm{\frac{e^t}{t}}$$\mathrm{d}t$ y

$e^t=\sum _{n=0}^\infty \frac{t^n}{n!}$

$\frac{e^t}{t}=\sum_{n=0}^\infty \frac{t^{n-1}}{n!}$

$\int \mathrm{\frac{e^t}{t}}\,\mathrm{d}t=\int \mathrm \sum{n=0}^\infty{\frac{t^{n-1}}{n!}}\,\mathrm{d}t=\sum{n=0}^\infty\frac{t^n}{n*(n)!}$

¿Cómo puedo solucionar $\int_0^1 \mathrm{\frac{x}{ln(x)}}$?

Gracias por su ayuda :) tenga un buen día

Answer 1

$\newcommand{\+}{^{\daga}}% \newcommand{\ángulos}[1]{\left\langle #1 \right\rangle}% \newcommand{\llaves}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\a la derecha\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \cima {= \cima \vphantom{\enorme}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\cy}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\int_{0}^{1}{x - 1 \over \ln\pars{x}}\,\dd x: {\large ?}.\quad}$ Con el cambio de las variables de $x \equiv \expo{-z}:$ \begin{align} \color{#0000ff}{\large\int_{0}^{1}{x - 1 \over \ln\pars{x}}\,\dd x} &= \int_{\infty}^{0}{\expo{-z} - 1 \over -z}\,\pars{-\expo{-z}\,\dd z} = \int^{\infty}_{0}{\expo{-z} - \expo{-2z}\over z}\,\dd z \\[3mm]&= -\int^{\infty}_{0}\ln\pars{z}\pars{-\expo{-z} +2 \expo{-2z}}\,\dd z = \int^{\infty}_{0}\ln\pars{z}\expo{-z}\,\dd z - \int^{\infty}_{0}\ln\pars{z \over 2}\expo{-z}\,\dd z \\[3mm]&= \int^{\infty}_{0}\ln\pars{z}\expo{-z}\,\dd z - \int^{\infty}_{0}\bracks{\ln\pars{z} - \ln\pars{2}}\expo{-z}\,\dd z = \ln\pars{2}\overbrace{\int^{\infty}_{0}\expo{-z}\,\dd z}^{=\ 1} \\[3mm]&= \color{#0000ff}{\large\ln\pars{2}} \end{align}