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Question Number 164231 by Zaynal last updated on 15/Jan/22 $$\int_{−\infty} ^{\infty} \:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\frac{\frac{\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{3}.^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{6}}{\:\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{9}.^{\mathrm{8}} \boldsymbol{\mathrm{log}}\:\mathrm{2}+\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{9}.^{\mathrm{8}} \boldsymbol{\mathrm{log}}\:\mathrm{3}}}{\frac{\:^{\mathrm{9}} \boldsymbol{\mathrm{log}}\:\mathrm{4}.^{\mathrm{2}} \boldsymbol{\mathrm{log}}\mathrm{27}+^{\mathrm{3}} \boldsymbol{\mathrm{log}}\:\mathrm{81}}{\:^{\mathrm{2}} \boldsymbol{\mathrm{log}}\:\mathrm{64}−^{\mathrm{2}} \boldsymbol{\mathrm{log}}\:\mathrm{8}}}\right)\boldsymbol{\mathrm{dxdy}}…
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Question Number 164211 by mnjuly1970 last updated on 15/Jan/22 Answered by smallEinstein last updated on 16/Jan/22 Commented by mr W last updated on 16/Jan/22 $${lordose}\:{sir}:\:…