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Question Number 145941 by Mathspace last updated on 09/Jul/21
find ∫_0 ^∞ e^(−3x) log(1+x^3 )dx
find∫0∞e−3xlog(1+x3)dx
Answered by ArielVyny last updated on 24/Jul/21
((3x^2 )/(1+x^3 ))=3Σ(−1)^n x^(3n+2) ln(1+x^3 )=Σ3(−1)^n (1/(3n+3))x^(3n+3) =Σ(((−1)^n )/(n+1))x^(3n+3) ∫_0 ^∞ e^(−3x) ln(1+x^3 )=Σ(((−1)^n )/(n+1))∫_0 ^∞ e^(−3x) x^(3n+3) dx 3x=t→3dx=dt ∫_0 ^∞ e^(−t) ((t/3))^(3n+3) (1/3)dt=((1/3))^(3n+4) ∫_0 ^∞ e^(−t) t^(3n+3) dt =((1/3))^(3n+4) Γ(3n+4)=((1/3))^(3n+4) (3n+3)! ∫_0 ^∞ e^(−3x) ln(1+x^3 )dx=Σ(((−1)^n )/(n+1))((1/3))^(3n+4) (3n+3)! =Σ(((−1)^n (3n+3))/(n+1))((1/3))^(3n+4) (3n+2)! =Σ(−1)^n 3×3^(−(3n+4)) (3n+2)! =Σ(−1)^n ((1/3))^(3n+3) (3n+2)! =(1/(27))Σ(−1)^n ((1/(27)))^n (3n+2)!
3x21+x3=3Σ(−1)nx3n+2ln(1+x3)=Σ3(−1)n13n+3x3n+3=Σ(−1)nn+1x3n+3∫0∞e−3xln(1+x3)=Σ(−1)nn+1∫0∞e−3xx3n+3dx3x=t→3dx=dt∫0∞e−t(t3)3n+313dt=(13)3n+4∫0∞e−tt3n+3dt=(13)3n+4Γ(3n+4)=(13)3n+4(3n+3)!∫0∞e−3xln(1+x3)dx=Σ(−1)nn+1(13)3n+4(3n+3)!=Σ(−1)n(3n+3)n+1(13)3n+4(3n+2)!=Σ(−1)n3×3−(3n+4)(3n+2)!=Σ(−1)n(13)3n+3(3n+2)!=127Σ(−1)n(127)n(3n+2)!

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