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 =???

Commented byMJS_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...

Commented byDwaipayan Shikari last updated on 18/Sep/20

((log(6))/3)

Commented byDwaipayan 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)

Commented byMJS_new last updated on 18/Sep/20

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

Commented bymnjuly1970 last updated on 18/Sep/20

grateful ..=

Commented bymnjuly1970 last updated on 18/Sep/20

thank you ....