Question #092c3

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Answer 1 · 2017-11-05T16:39:34

$int_0^1 x^3*[Ln(1/x)]^3*dx$ = $3/128$

$int_0^1 x^3*[Ln(1/x)]^3*dx$

After using $u=Ln(1/x)=-Lnx$ , $Lnx=-u$ , $x=e^(-u)$ and $dx=-e^(-u)*du$ transforms, this integral became,

$int_oo^0 -u^3*e^(-4u)*du$

$=int_0^oo u^3*e^(-4u)*du$

After using tabular integration,

$=int_0^oo u^3*e^(-4u)*du$

= $[u^3*(-1/4e^(-4u))-3u^2*1/16e^(-4u)+6u*(-1/64*e^(-4u))-6*1/256e^(-4u)]_0^oo$

= $[-1/128e^(-4u)*(32u^3+24u^2+12u+3)]_0^oo$

= $3/128$