# Indefinite Integrals

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find the indefinite integral of

(1/x)*(e^(-2log[ 3,x])) dx

where 3 is base of the log

• Indefinite Integrals -

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)

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