Evaluating integral with $\sqrt[3] x$ and $\ln x$ I found this example amongst set of examples to get ready on exam
$$
\int\frac{\ln x}{\large\sqrt[3]{x}}\ dx
$$
I am able to see the basic substitution, but I don't know how to count if further… Anyone who could help me with this one?
Thank you.
 A: Hint
Use integration by parts with $u=\log(x)$ and $v'=x^{-1/3}~dx$. So, $u'=\frac {dx}{x}$ and $v=\frac{3 x^{2/3}}{2}$.
I am sure that you can take from here.
A: Using IBP, let $u=\ln x$, $du=\dfrac1x\ dx$, $dv=x^{-\Large\frac13}\ dx$, and
$$
v=\int x^{-\Large\frac13}\ dx=\frac32x^{\Large\frac23},
$$
then
$$
\int\frac{\ln x}{\sqrt[3]{x}}\ dx=\frac32x^{\Large\frac23}\ln x-\int\frac32x^{\Large\frac23}\dfrac1x\ dx=\frac32x^{\Large\frac23}\ln x-\frac32\int x^{-\Large\frac13}\ dx.
$$
I leave it the rest for you.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\int x^{\mu}\,\dd x&={x^{\mu + 1} \over \mu + 1}\ \imp\
\int x^{\mu}\ln\pars{x}\,\dd x=-\,{x^{1 + \mu} \over \pars{1 + \mu}^{2}}
+ {x^{1 + \mu}\ln\pars{x} \over 1 + \mu}
\end{align}

Set $\ds{\mu = -\,{1 \over 3}}$:
  $$\color{#00f}{\large%
\int {\ln\pars{x} \over \root[3]{x}}\,\dd x=-\,{9x^{1/3} \over 4}
+ {3x^{2/3}\ln\pars{x} \over 2}} + \mbox{a constant}
$$

