How do I integrate this natural logarithmic function?
$$ \int 5 \mathrm{ln}(x^{1/3}) dx$$ Working with natural log is very unfamiliar to me (I think I was cheated in my Calc 1 course). I know that integrating $\frac{1}{x} = \ln x$, but is it the same in reverse? Also would this simply be $u$ - substitution or instead integral by parts?
Help is so very appreciated!
$\endgroup$ 12 Answers
$\begingroup$First, clean it up a bit using some properties of logarithms:
$$5 \ln{x^{1/3}} = 5 \left(\frac 1 3 \ln{x}\right) = \frac 5 3 \ln{x}$$
Now in order to integrate $\ln$, try going by parts with $u = \ln{x}$ and $dv = dx$. This gives
$$\int \ln{x} dx = x \ln{x} - \int x \frac{dx}{x}$$
$\endgroup$ 4 $\begingroup$You can also substitute away $\mathrm{ln}x$ altogether. You will then end up with the well known integral $xe^x$, kind of the introductory example of Integration by Parts.
Considering your earlier post about $x\mathrm{cos}(6x)$, I may have the inclination that you have seen this integral before. Back-substitute gives the anti-derivative from Bongers.
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