6

$$I=\int_{0}^{1}{6x(x-1)(x+2)\over (x+1)^3}\ln(x)dx=(\pi-3)(\pi+3)\tag1$$

$$I=\int_{0}^{1}\left(6-{12\over 1+x}-{6\over (1+x)^2}+{12\over (1+x)^3}\right)\ln(x)dx\tag2$$

Recall

$$\int_{0}^{1}{\ln(x)\over 1+x}dx=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}x^n\ln(x)dx=-\sum_{n=0}^{\infty}{(-1)^n\over (n+1)^2}=-{\pi^2\over 12}$$

$$\int_{0}^{1}{\ln(x)\over (1+x)^2}dx=\sum_{n=0}^{\infty}(-1)^n(n+1)\int_{0}^{1}x^n\ln(x)dx=-\sum_{n=0}^{\infty}{(-1)^n(n+1)\over (n+1)^2}=-\ln(2)$$

see answer of felix marin

Substitute into (2)$\rightarrow$ (3)

$$I=-6+\pi^2+6\ln(2)-12\cdot{1\over 4}[2\ln(2)+1]\tag3$$

$$I=\pi^2-9=(\pi-3)(\pi+3)\tag4$$

Anyone can prove I using an another approach? (Prefer quicker technique)

Community
  • 1
gymbvghjkgkjkhgfkl
  • 8,090

3 Answers3

5

Hint. One may integrate by parts. $$ \begin{align} &\int_{0}^{1}{6x(x-1)(x+2)\over (x+1)^3}\ln(x)dx \\\\&=\left. \left(6x+\frac{6x}{(1+x)^2}-12 \ln(1+x)\right)\ln x\right|_0^1-\int_0^1\left(6+\frac6{(1+x)^2}-12\frac{\ln(1+x)}{x}\right)dx \\\\&=0-(9-\pi ^2) \\\\&=(\pi-3)(\pi+3) \end{align} $$ where we have used the standard result $$ 12\int_0^1\frac{\ln(1+x)}{x}dx=-12 \:\text{Li}_2(-1)=\pi^2. $$

Olivier Oloa
  • 120,989
2

An alternative approach. Since: $$\forall x\in(0,1),\qquad \frac{x(x-1)(x+2)}{(x+1)^3}=\sum_{n\geq 1}(-1)^n (n^2+2n-1)x^n \tag{1}$$ we have: $$\int_{0}^{1}\frac{x(x-1)(x+2)}{(x+1)^3}\,\log(x)\,dx = \sum_{n\geq 1}'(-1)^{n+1}\frac{n^2+2n-1}{(n+1)^2}=\frac{3}{2}-2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}\tag{2} $$ where $\sum'$ has to be intended à-la-Cesàro/Abel/Borel: $\sum_{n\geq 1}'a_n = \lim_{x\to 0^+}\sum_{n\geq 1}a_n e^{-nx}.$

In the RHS of $(2)$ we may easily recognize $\eta(2)=\frac{\zeta(2)}{2}=\frac{\pi^2}{12}$ and the claim easily follows.

Jack D'Aurizio
  • 353,855
1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{I} & = \int_{0}^{1}{6x\pars{x - 1}\pars{x + 2} \over \pars{x + 1}^{3}}\ln\pars{x} \,\dd x \\[4mm] & = 12\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}^{3}}\,\dd x - 6\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x - 12\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x + 6\ \overbrace{\int_{0}^{1}\ln\pars{x}\,\dd x}^{\ds{-1}} \end{align}

\begin{align} \fbox{$\ds{\ \int_{0}^{1}{\ln\pars{x} \over a + x}\,\dd x\ }$} & = -\int_{0}^{1}{\ln\pars{x} \over 1 - x/\pars{-a}}\,{\dd x \over -a} = -\int_{0}^{-1/a}\,{\ln\pars{-ax} \over 1 - x}\,\dd x \\[4mm] & = -\int_{0}^{-1/a}{\ln\pars{1 - x} \over x}\,\dd x = \fbox{$\ds{\ \Li{2}\pars{-\,{1 \over a}}\ }$} \end{align}
$$ \left\lbrace\begin{array}{rcccl} \ds{\int_{0}^{1}{\ln\pars{x} \over x + 1}} & \ds{=} & \ds{\Li{2}\pars{-1}} & \ds{=} & \ds{-\,{\pi^{2} \over 12}} \\[3mm] \ds{\int_{0}^{1}{\ln\pars{x} \over \pars{x + 1}^{2}}} & \ds{=} & \ds{\left.-\,\totald{\Li{2}\pars{-1/a}}{a}\right\vert_{\ a\ =\ 1}} & \ds{=} & \ds{-\ln\pars{2}} \\[3mm] \ds{\int_{0}^{1}{\ln\pars{x} \over \pars{x + 1}^{3}}} & \ds{=} & \ds{\left.\half\,\totald[2]{\Li{2}\pars{-1/a}}{a}\right\vert_{\ a\ =\ 1}} & \ds{=} & \ds{-\,{1 \over 4} - \half\,\ln\pars{2}} \end{array}\right. $$
\begin{align} \color{#f00}{I} & = \int_{0}^{1}{6x\pars{x - 1}\pars{x + 2} \over \pars{x + 1}^{3}}\ln\pars{x} \,\dd x \\[4mm] & = 12\bracks{-\,{1 \over 4} - \half\,\ln\pars{2}} - 6\bracks{-\ln\pars{2}} - 12\pars{-\,{\pi^{2} \over 12}} -6 = \pi^{2} - 9 = \color{#f00}{\pars{\pi - 3}\pars{\pi + 3}} \end{align}
Felix Marin
  • 89,464