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Find the value of

$$\int_0^1(x \ln x)^3dx $$

Taking a substitute $x=e^{-y}$ i get the value as $$-3\over128$$

Does it look good ?

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  • $\begingroup$ yeah that's right. $\endgroup$
    – Rustyn
    Commented May 2, 2013 at 15:30
  • $\begingroup$ @Aman : have you used wolframalpha.com ? You can check your answer there. They seem to accept any reasonble syntax, such as Maplese: int((x*ln(x))^3),x=0..1) would probably work. $\endgroup$ Commented May 2, 2013 at 15:42
  • $\begingroup$ @Aman: btw, I don't think it's really an improper integral, because $(x\ln x)^3 \to 0$ as $x \to 0^+$. $\endgroup$ Commented May 2, 2013 at 15:50
  • $\begingroup$ It is improper in the sense that x=0 is not in the domain of the given function. $\endgroup$
    – imranfat
    Commented May 2, 2013 at 17:29

1 Answer 1

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Mathematica returns your answer too:

In:= Integrate[(x Log[x])^3,{x,0,1}]

Out=-3/128

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