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Question Number 114304 by mnjuly1970 last updated on 18/Sep/20

       calculus.....   evaluate:                Ω=∫_(−1) ^( 1) xln(1^x +2^x +3^x +6^x )dx =???

$$\:\:\:\:\:\:\:{calculus}..... \\ $$$$\:{evaluate}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{−\mathrm{1}} ^{\:\mathrm{1}} {xln}\left(\mathrm{1}^{{x}} +\mathrm{2}^{{x}} +\mathrm{3}^{{x}} +\mathrm{6}^{{x}} \right){dx}\:=???\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Commented by MJS_new last updated on 18/Sep/20

=∫_(−1) ^1 xln ((1+2^x )(1+3^x )) dx=  =∫_(−1) ^1 xln (1+2^x ) dx+∫_(−1) ^1 xln (1+3^x ) dx  looks easier now...

$$=\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}{x}\mathrm{ln}\:\left(\left(\mathrm{1}+\mathrm{2}^{{x}} \right)\left(\mathrm{1}+\mathrm{3}^{{x}} \right)\right)\:{dx}= \\ $$$$=\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}{x}\mathrm{ln}\:\left(\mathrm{1}+\mathrm{2}^{{x}} \right)\:{dx}+\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}{x}\mathrm{ln}\:\left(\mathrm{1}+\mathrm{3}^{{x}} \right)\:{dx} \\ $$$$\mathrm{looks}\:\mathrm{easier}\:\mathrm{now}... \\ $$

Commented by Dwaipayan Shikari last updated on 18/Sep/20

((log(6))/3)

$$\frac{{log}\left(\mathrm{6}\right)}{\mathrm{3}} \\ $$

Commented by Dwaipayan Shikari last updated on 18/Sep/20

∫_(−1) ^1 xlog(1+2^x )+xlog(1+3^x )dx  =I_a +I_b   I_a =∫_(−1) ^1 xlog(1+2^x )=∫_(−1) ^1 −xlog(1+2^x )+xlog2^x =I_a   2I_a =2∫_0 ^1 x^2 log(2)  I_a =((log(2))/3)  I_b =((log(3))/3)  I_a +I_b =((log(6))/3)

$$\int_{−\mathrm{1}} ^{\mathrm{1}} {xlog}\left(\mathrm{1}+\mathrm{2}^{{x}} \right)+{xlog}\left(\mathrm{1}+\mathrm{3}^{{x}} \right){dx} \\ $$$$={I}_{{a}} +{I}_{{b}} \\ $$$${I}_{{a}} =\int_{−\mathrm{1}} ^{\mathrm{1}} {xlog}\left(\mathrm{1}+\mathrm{2}^{{x}} \right)=\int_{−\mathrm{1}} ^{\mathrm{1}} −{xlog}\left(\mathrm{1}+\mathrm{2}^{{x}} \right)+{xlog}\mathrm{2}^{{x}} ={I}_{{a}} \\ $$$$\mathrm{2}{I}_{{a}} =\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {log}\left(\mathrm{2}\right) \\ $$$${I}_{{a}} =\frac{{log}\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${I}_{{b}} =\frac{{log}\left(\mathrm{3}\right)}{\mathrm{3}} \\ $$$${I}_{{a}} +{I}_{{b}} =\frac{{log}\left(\mathrm{6}\right)}{\mathrm{3}} \\ $$

Commented by MJS_new last updated on 18/Sep/20

yesss! I had been focused on the integral,  forgot the symmetric borders...

$$\mathrm{yesss}!\:\mathrm{I}\:\mathrm{had}\:\mathrm{been}\:\mathrm{focused}\:\mathrm{on}\:\mathrm{the}\:\mathrm{integral}, \\ $$$$\mathrm{forgot}\:\mathrm{the}\:\mathrm{symmetric}\:\mathrm{borders}... \\ $$

Commented by mnjuly1970 last updated on 18/Sep/20

grateful ..=

$${grateful}\:..= \\ $$

Commented by mnjuly1970 last updated on 18/Sep/20

thank you ....

$${thank}\:{you}\:.... \\ $$

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