integrate by parts: ʃ sinx lnx
I don't think this is expressible in terms of elementary functions
can you cycle it back around?
sin(x) + ln(x) cos(x) - 1/x -sin(x) hmmm
the way I see it: dv=sin x u=lnx u never goes away the other way, u=lnx, does not seem promising, but give it a try :D
the other way dv=lnx I mean...
ln(x) ups to what: xln(x) - x i think
lnx cant be integrated right?
yes, lnx can be integrated
xlnx - x derives to ln(x) + x/x - 1 = ln(x)
\[\int sin(x)ln(x)dx= sin(x)(xln(x)-x) - \int cos(x)(xlnx-x) dx\] \[\int cos(x)(xlnx-x) dx=\int xcos(x)ln(x)-xcos(x)\ dx\]
its ugly, but might be doable overall
lnx cant be differentiated?
otherwise: \[\int sin(x)ln(x)dx=cos(x)ln(x)-\int \frac{cos(x)}{x}dx\]
really? ln(x)' = 1/x
you need to brush up on this stuff i assume lol
lnx cant be integrated im sure. it is non-integrable
lnx is totally integrable by parts, as amistre did above
\[\int \ln xdx\]\[u=\ln x\]\[dv=1\]try it!
differntial of lnx=1/x.... but what is the answer of integral of lnx? just the final answer?
um... the final answer is the final answer, not sure what you mean
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