Question Number 164231 by Zaynal last updated on 15/Jan/22
$$\int_{−\infty} ^{\infty} \:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\frac{\frac{\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{3}.^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{6}}{\:\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{9}.^{\mathrm{8}} \boldsymbol{\mathrm{log}}\:\mathrm{2}+\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{9}.^{\mathrm{8}} \boldsymbol{\mathrm{log}}\:\mathrm{3}}}{\frac{\:^{\mathrm{9}} \boldsymbol{\mathrm{log}}\:\mathrm{4}.^{\mathrm{2}} \boldsymbol{\mathrm{log}}\mathrm{27}+^{\mathrm{3}} \boldsymbol{\mathrm{log}}\:\mathrm{81}}{\:^{\mathrm{2}} \boldsymbol{\mathrm{log}}\:\mathrm{64}−^{\mathrm{2}} \boldsymbol{\mathrm{log}}\:\mathrm{8}}}\right)\boldsymbol{\mathrm{dxdy}} \\ $$$$\left\{\boldsymbol{\mathrm{z}}.\right\} \\ $$