Evaluate: ${{\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx}}$

Question

Evaluate: $${{I=\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx.}}$$

The answer is given below: $$ I=-\frac{7}{12}\pi^4\ln^2(2)-\frac{35}{12}\pi^2\ln^4(2) +\frac{13}{2}\ln^6(2)+\ln^5(2)\ln(3)- 5\ln^4(2)\operatorname{Li}_2\left ( -\frac{1}{2} \right ) -5\ln^3(2)\operatorname{Li}_3\left (\frac{1}{4} \right ) -60\ln^2(2)\operatorname{Li}_4\left ( -\frac{1}{2} \right ) +\frac{95}{2}\ln^3(2)\zeta(3)\approx0.0553825. $$ How to prove this?

Answer 1

$$\left( \mathrm{I} \right) \\ I=\int_0^1{\frac{\ln ^5\left( 1+x \right)}{2+x}dx}-\int_0^1{\frac{\ln ^5\left( 1+x \right)}{3+x}dx}+5\ln 2\int_0^1{\frac{\ln ^4\left( 1+x \right)}{3+x}dx} \\ =\int_1^2{\frac{\ln ^5x}{1+x}dx}-\int_1^2{\frac{\ln ^5x}{2+x}dx}+5\ln 2\int_1^2{\frac{\ln ^4x}{2+x}dx} \\ =-\int_{\frac{1}{2}}^1{\frac{\ln ^5x}{1+x}\frac{1}{x}dx}-\int_{\frac{1}{2}}^1{\frac{\ln ^5\left( 2t \right)}{1+t}dt}+5\ln 2\int_{\frac{1}{2}}^1{\frac{\ln ^4\left( 2t \right)}{1+t}dt} \\ =\int_{\frac{1}{2}}^1{\frac{\ln ^5x}{1+x}dx}-\int_{\frac{1}{2}}^1{\frac{\ln ^5x}{x}dx}-\int_{\frac{1}{2}}^1{\frac{\ln ^5\left( 2t \right)}{1+t}dt}+5\ln 2\int_{\frac{1}{2}}^1{\frac{\ln ^4\left( 2t \right)}{1+t}dt} \\ =\int_{\frac{1}{2}}^1{\frac{\ln ^5x}{1+x}dx}+\frac{1}{6}\ln ^62-\int_{\frac{1}{2}}^1{\frac{\ln ^5\left( 2t \right)}{1+t}dt}+5\ln 2\int_{\frac{1}{2}}^1{\frac{\ln ^4\left( 2t \right)}{1+t}dt} \\ =\int_0^1{\frac{\ln ^5x}{1+x}dx}-\int_0^{\frac{1}{2}}{\frac{\ln ^5x}{1+x}dx}+\frac{1}{6}\ln ^62-\int_0^1{\frac{\ln ^5\left( 2t \right)}{1+t}dt}+\int_0^{\frac{1}{2}}{\frac{\ln ^5\left( 2t \right)}{1+t}dt}+5\ln 2\int_0^1{\frac{\ln ^4\left( 2t \right)}{1+t}dt}-5\ln 2\int_0^{\frac{1}{2}}{\frac{\ln ^4\left( 2t \right)}{1+t}dt} \\ =\int_0^1{\frac{\ln ^5x}{1+x}dx}-\int_0^1{\frac{\frac{1}{2}\ln ^5\frac{x}{2}}{1+\frac{1}{2}x}dx}+\frac{1}{6}\ln ^62-\int_0^1{\frac{\ln ^5\left( 2t \right)}{1+t}dt}+\int_0^1{\frac{\frac{1}{2}\ln ^5u}{1+\frac{1}{2}u}du}+5\ln 2\int_0^1{\frac{\ln ^4\left( 2t \right)}{1+t}dt}-5\ln 2\int_0^1{\frac{\frac{1}{2}\ln ^4u}{1+\frac{1}{2}u}du} \\ =\int_0^1{\frac{\ln ^5x}{1+x}dx}-\int_0^1{\frac{\frac{1}{2}\ln ^5x}{1+\frac{1}{2}x}dx}+5\ln 2\int_0^1{\frac{\frac{1}{2}\ln ^4x}{1+\frac{1}{2}x}dx}-10\ln ^22\int_0^1{\frac{\frac{1}{2}\ln ^3x}{1+\frac{1}{2}x}dx}+10\ln ^32\int_0^1{\frac{\frac{1}{2}\ln ^2x}{1+\frac{1}{2}x}dx} \\ -5\ln ^42\int_0^1{\frac{\frac{1}{2}\ln x}{1+\frac{1}{2}x}dx}+\ln ^52\int_0^1{\frac{\frac{1}{2}}{1+\frac{1}{2}x}dx}+\frac{1}{6}\ln ^62-\int_0^1{\frac{\ln ^5t}{1+t}dt}-5\ln 2\int_0^1{\frac{\ln ^4t}{1+t}dt}-10\ln ^22\int_0^1{\frac{\ln ^3t}{1+t}dt} \\ -10\ln ^32\int_0^1{\frac{\ln ^2t}{1+t}dt}-5\ln ^42\int_0^1{\frac{\ln t}{1+t}dt}-\ln ^52\int_0^1{\frac{1}{1+t}dt}+\int_0^1{\frac{\frac{1}{2}\ln ^5u}{1+\frac{1}{2}u}du}+5\ln 2\int_0^1{\frac{\ln ^4t}{1+t}dt} \\ +20\ln ^22\int_0^1{\frac{\ln ^3t}{1+t}dt}+30\ln ^32\int_0^1{\frac{\ln ^2t}{1+t}dt}+20\ln ^42\int_0^1{\frac{\ln t}{1+t}dt}+5\ln ^52\int_0^1{\frac{1}{1+t}dt}-5\ln 2\int_0^1{\frac{\frac{1}{2}\ln ^4u}{1+\frac{1}{2}u}du} \\ =-10\ln ^22\int_0^1{\frac{\frac{1}{2}\ln ^3x}{1+\frac{1}{2}x}dx}+10\ln ^32\int_0^1{\frac{\frac{1}{2}\ln ^2x}{1+\frac{1}{2}x}dx}-5\ln ^42\int_0^1{\frac{\frac{1}{2}\ln x}{1+\frac{1}{2}x}dx}+\ln ^52\int_0^1{\frac{\frac{1}{2}}{1+\frac{1}{2}x}dx} \\ +\frac{25}{6}\ln ^62+10\ln ^22\int_0^1{\frac{\ln ^3t}{1+t}dt}+20\ln ^32\int_0^1{\frac{\ln ^2t}{1+t}dt}+15\ln ^42\int_0^1{\frac{\ln t}{1+t}dt} \\ =-60\ln ^22\mathrm{Li}_4\left( -\frac{1}{2} \right) -20\ln ^32\mathrm{Li}_3\left( -\frac{1}{2} \right) -5\ln ^42\mathrm{Li}_2\left( -\frac{1}{2} \right) +\ln ^52\ln 3 \\ +\frac{19}{6}\ln ^62-\frac{7}{12}\pi ^4\ln ^22+30\zeta \left( 3 \right) \ln ^32-\frac{5}{4}\pi ^2\ln ^42 \\ \left( \mathrm{II} \right) \\ \mathrm{Li}_3\left( -\frac{1}{2} \right) =-\frac{1}{2}\int_0^1{\frac{\frac{1}{2}\ln ^2x}{1+\frac{1}{2}x}dx} \\ \mathrm{Li}_3\left( \frac{1}{4} \right) =\frac{1}{2}\int_0^1{\frac{\frac{1}{4}\ln ^2x}{1-\frac{1}{4}x}dx}=\frac{1}{2}\int_0^1{\frac{\ln ^2x}{4-x}dx}=4\int_0^1{\frac{t\ln ^2t}{4-t^2}dt} \\ =2\int_0^1{\frac{\ln ^2t}{2-t}dt}-2\int_0^1{\frac{\ln ^2t}{2+t}dt} \\ =4\mathrm{Li}_3\left( \frac{1}{2} \right) +4\mathrm{Li}_3\left( -\frac{1}{2} \right) \\ \mathrm{Li}_3\left( -\frac{1}{2} \right) =\frac{\pi ^2}{12}\ln 2-\frac{7}{8}\zeta \left( 3 \right) -\frac{1}{6}\ln ^32+\frac{1}{4}\mathrm{Li}_3\left( \frac{1}{4} \right) \\ \left( \mathrm{III} \right) \\ I=-60\ln ^22\mathrm{Li}_4\left( -\frac{1}{2} \right) -20\ln ^32\left( \frac{\pi ^2}{12}\ln 2-\frac{7}{8}\zeta \left( 3 \right) -\frac{1}{6}\ln ^32+\frac{1}{4}\mathrm{Li}_3\left( \frac{1}{4} \right) \right) -5\ln ^42\mathrm{Li}_2\left( -\frac{1}{2} \right) +\ln ^52\ln 3 \\ +\frac{19}{6}\ln ^62-\frac{7}{12}\pi ^4\ln ^22+30\zeta \left( 3 \right) \ln ^32-\frac{5}{4}\pi ^2\ln ^42 \\ =-60\ln ^22\mathrm{Li}_4\left( -\frac{1}{2} \right) -5\ln ^42\mathrm{Li}_2\left( -\frac{1}{2} \right) +\ln ^52\ln 3+\frac{13}{2}\ln ^62 \\ -\frac{7}{12}\pi ^4\ln ^22+\frac{95}{2}\zeta \left( 3 \right) \ln ^32-\frac{35}{12}\pi ^2\ln ^42-5\ln ^32\mathrm{Li}_3\left( \frac{1}{4} \right) $$