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indefinite integral of 1/xln(x^3) dx
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I have work so far but as usual it is probably wrong...
kind of....
here's a hint.. let \[u = \ln (x^3)\] so \[ du = \frac 3xdx\] agree?
Yeah.. so one gets integral 1/u*(3/xdx)
and then do you move the 3/x to the outside?
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hmm wait.. im getting confused with these x and u's combined
\[du = \frac 3x dx\] so \[\frac{du}{3} = \frac{dx}{x}\] \[\int \frac{dx}{x \ln x^3} \implies \int \frac{du}{3u}\] right?
Oh, okay... yeah
so take out 1/3 out of the integral \[\implies \frac 13\int \frac{du}{u}\] now this is basic integration right?
right because integral u/du is the same as integral 1/u du which is log u
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sub for u and multiply by 1/3?
right.
meant du/u* okay thanks
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