# int (ln^3 x)/(x) dx Is this the same as int ((ln x)^3)/(x)dx?

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Is this the same as
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Unfortunately, "shorthand" can lead to ambiguity:
${\mathrm{ln}}^{3}x=\left(\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{ln}x\right)\right)\phantom{\rule{thickmathspace}{0ex}}?\phantom{\rule{1em}{0ex}}$ or $\phantom{\rule{1em}{0ex}}{\mathrm{ln}}^{3}x=\left(\mathrm{ln}x{\right)}^{3}\phantom{\rule{thickmathspace}{0ex}}?$
But as you suspected, in this context, and given the integral, I'm am quite sure that ${\mathrm{ln}}^{3}x=\left(\mathrm{ln}x{\right)}^{3},\phantom{\rule{thickmathspace}{0ex}}$ much like $\phantom{\rule{thickmathspace}{0ex}}{\mathrm{sin}}^{2}\left(x\right)=\left(\mathrm{sin}x{\right)}^{2}.\phantom{\rule{thickmathspace}{0ex}}$ So your integral amounts to:

Let $u=\mathrm{ln}\left(x\right),;\phantom{\rule{thickmathspace}{0ex}}du=\frac{dx}{x}$
$\int \frac{\left(\mathrm{ln}x{\right)}^{3}}{x}\phantom{\rule{thinmathspace}{0ex}}dx=\int {u}^{3}\phantom{\rule{thinmathspace}{0ex}}du=\frac{1}{4}{u}^{4}+C$
$=\frac{1}{4}\left(\mathrm{ln}x{\right)}^{4}+C=\frac{1}{4}{\mathrm{ln}}^{4}x+C$