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find integral of {ln(x)^3/x dx
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let u=ln(x)
provided you mean \[[ \ln(x)]^3 / x\]
u=ln(x) du= (1/x) dx integral = u^3 du = u^4 / 4 +C = [ ln(x)]^4 +C
Just as elec said :) let u = ln(x) then du = 1/x and you'll get the following: \[=\int\limits u^3 du \]\[= \frac{u^4}{4} + c \rightarrow = \frac{(\ln|x|)^4}{4} + c\]
EDIT: (1/4) [ ln(x)]^4 +C
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LOL! :)
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