Asked by Erika
find the indefinite integral of
(1/x)*(e^(-2log[ 3,x])) dx
where 3 is base of the log
i cannot even figure out where to start :( please help. thank you
(1/x)*(e^(-2log[ 3,x])) dx
where 3 is base of the log
i cannot even figure out where to start :( please help. thank you
Answers
Answered by
Steve
log_3(x) = lnx/ln3
e^(-2log_3(x))
= e^(-2lnx/ln3)
= (e^(lnx))^(-2/ln3)
= x^(-2/ln3)
so you have
∫(1/x) x^(-2/ln3) dx
= ∫x^(-2/ln3-1) dx
= 1/(-2/ln3) x^(-2/ln3)
or
-ln3/2 x^(-2/ln3)
e^(-2log_3(x))
= e^(-2lnx/ln3)
= (e^(lnx))^(-2/ln3)
= x^(-2/ln3)
so you have
∫(1/x) x^(-2/ln3) dx
= ∫x^(-2/ln3-1) dx
= 1/(-2/ln3) x^(-2/ln3)
or
-ln3/2 x^(-2/ln3)
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