calculus-evaluate-1-1-xln-1-x-2-x-3-x-6-x-dx-

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

c a l c u l u s \dots . .

e v a l u a t e :

Ω = \int_{- 1}^{1} x l n (1^{x} + 2^{x} + 3^{x} + 6^{x}) d x = ? ? ?

Commented by MJS_new last updated on 18/Sep/20

= \underset{- 1}{\int^{1}} x \ln ((1 + 2^{x}) (1 + 3^{x})) d x =

= \underset{- 1}{\int^{1}} x \ln (1 + 2^{x}) d x + \underset{- 1}{\int^{1}} x \ln (1 + 3^{x}) d x

looks easier now \dots

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

\frac{l o g (6)}{3}

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

\int_{- 1}^{1} x l o g (1 + 2^{x}) + x l o g (1 + 3^{x}) d x

= I_{a} + I_{b}

I_{a} = \int_{- 1}^{1} x l o g (1 + 2^{x}) = \int_{- 1}^{1} - x l o g (1 + 2^{x}) + x l o g 2^{x} = I_{a}

2 I_{a} = 2 \int_{0}^{1} x^{2} l o g (2)

I_{a} = \frac{l o g (2)}{3}

I_{b} = \frac{l o g (3)}{3}

I_{a} + I_{b} = \frac{l o g (6)}{3}

Commented by MJS_new last updated on 18/Sep/20

yesss! I had been focused on the integral,

forgot the symmetric borders \dots

Commented by mnjuly1970 last updated on 18/Sep/20

g r a t e f u l . . =

Commented by mnjuly1970 last updated on 18/Sep/20

t h a n k y o u \dots .