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Question Number 142426 by Mathspace last updated on 31/May/21

find the value of ∫_0 ^∞ ((xlogx)/((1+x^3 )^2 ))dx

findthevalueof∫0∞xlogx(1+x3)2dx

Answered by mindispower last updated on 01/Jun/21

x→(1/x) =−∫_0 ^∞ ((x^3 ln(x)dx)/((1+x^3 )^2 )) =−[−((xln(x))/(3(1+x^3 )))]_0 ^∞ +(1/3)∫_0 ^∞ (1/(1+x^3 ))dx+(1/3)∫_0 ^∞ ((ln(x))/(1+x^3 ))dx] −(1/9)∫_0 ^∞ (t^((−2)/3) /(1+t))dt−(1/(27))∫_0 ^∞ ((t^(−(2/3)) ln(t))/(1+t))dt β(x,y)=∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) )) −(1/9)β((1/3),(2/3))−(1/(27))β_x ((1/3),(2/3)) =−((2π)/(9(√3)))−(1/(27))β((1/3),(2/3))(Ψ(1)−Ψ((1/3))) =−((2π)/(9(√3)))(1+(1/3)(−γ+γ+(π/(2(√3)))+(3/2)ln(3)) =−((2π)/(9(√3)))−(π^2 /(81))−((πln(3))/(9(√3)))

x→1x=−∫0∞x3ln(x)dx(1+x3)2=−[−xln(x)3(1+x3)]0∞+13∫0∞11+x3dx+13∫0∞ln(x)1+x3dx]−19∫0∞t−231+tdt−127∫0∞t−23ln(t)1+tdtβ(x,y)=∫0∞tx−1(1+t)x+y−19β(13,23)−127βx(13,23)=−2π93−127β(13,23)(Ψ(1)−Ψ(13))=−2π93(1+13(−γ+γ+π23+32ln(3))=−2π93−π281−πln(3)93

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