`L e tA=int_0^oo(logx)/(1+x^3)dxdotT h e nfin dt h ev a l u eofint_0^oo(xlogx)/(1+x^3)dx` in terms of `Adot`

Explanation

`B=int_(0)^(oo)(x log x)/(1+x^(3))dx`
`=int_(0)^(oo) ((x+1)logx-log)/(1+x^(3))dx`
`=int_(0)^(oo) (logx)/(x^(2)-x+1)dx-A`
let `I=int_(0)^(oo) (logx)/(x^(2)-x+1) dx`
Put `x=1/t`
`:. I=int_(oo)^(0) ("log"1/t)/(-1/(t^(2))-1/t+1)((-dt)/(t^(2)))`
`=-int_(0)^(oo) (logt)/(t^(2)-t+1)dt`
`=-I`
or `2I=0`
or `I=0`
`:.B=-A`