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July 29, 2014

July 29, 2014

Posted by **Erika** on Friday, May 24, 2013 at 9:55am.

(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

- Indefinite Integrals -
**Steve**, Friday, May 24, 2013 at 10:38amlog_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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