integral of $ \int (x^3+1)^{1/3} / x^2 dx $ What do you think about 
$$ \int \frac {(x^3+1)^{1/3}}{x^2}\,  dx $$
how do you compute this ? Is it possible to use the Euler substitution?
In fact, I don't know if the integral on my sheet is $$ \int \frac {(x^3+1)^{1/3}}  {x^2} \, dx $$
or 
$$ \int \frac{x^2 + 1}{x \sqrt {x^4 - x^2 + 1} } \frac {(x^3+1)^{1/3}}  {x^2}  \,dx $$
would the second expression make more sense? I thought that the second one is a little bit over complicated.
 A: Hint :  Divide numerator & denominator by $x$ & write the integrand  as $I= \int\frac {(1+\frac1 {x^3})^{1/3}}x dx $
Now, write $y= (1/x)^3 \implies  I = (-1/3)\int \frac {(1+y)^{1/3}}ydy $
Now, write $1+y =z^3 \implies dy=3z^2dz \implies  I = \int \frac {z^3}{1-z^3}dz$ 
From here you can use partial fraction to calculate the final integral.  
A: 
Hint :  Divide numerator & denominator by $x$ & write the integrand 
  as $I= \int\frac {(1+\frac1 {x^3})^{1/3}}x dx $

The whole calculus :
\begin{align}
 I  =& \int \frac{(x^3 +1)^{1/3} }{x^2} dx \\
    =& \int \frac{(1+ x^{-3})^{1/3} }{x} dx
\end{align}

let $y= x^{-3} \implies dy = \frac{-1}{3} x^{-4} dx $ 
$$ I = \frac{-1}{3} \int \frac {(1+y)^{1/3}}ydy  $$

let $ z^3 = 1+y \implies  dz =3z^2dy$ 
\begin{align}
 I &= \int \frac {z^3}{1-z^3}dz \\
   &= \int \frac{z^3 + 1 - 1}{1-z^3}dz \\
   &= \int \frac{z^3 - 1}{1-z^3}dz + \int \frac{ 1 }{1-z^3}dz \\
   &= - \int  dz + \int \frac{ 1 }{1-z^3}dz \\
however \\
   \int  \frac{ 1 }{z^3-1}dz  &= \int \frac{1}{  (z-1)(z^2+z+1) } dz \\ 
   &= \frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{3} \int \frac{z+2}{z^2 + z + 1} dz  \\ 
   &=\frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{6} \int \frac{2z+1}{z^2 + z + 1} dz  - \frac{1}{2} \int \frac{1}{z^2 + z + 1} dz \\
   &=\frac{1}{3} \int \frac{1}{z-1} dz - \frac{1}{6} \int \frac{2z+1}{z^2 + z + 1} dz  - \frac{1}{2} \int \frac{1}{ (z+ \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2 } dz \\
   &=\frac{1}{3} \ln( z-1 ) - \frac{1}{6} \ln(z^2 + z + 1) - \frac{1}{\sqrt 3} \arctan ( \frac{2z + 1}{ \sqrt 3} ) + cst \\
so \\
I &= -z + \frac{1}{3} \ln( z-1 ) - \frac{1}{6} \ln(z^2 + z + 1) - \frac{1}{\sqrt 3} \arctan ( \frac{2z + 1}{ \sqrt 3} ) + cst \\
i.e. \\
I &= -\sqrt[3]{1 + x^{-3} } + \frac{1}{3} \ln( \sqrt[3]{1 + x^{-3} } -1 ) - \frac{1}{6} \ln(\sqrt[3]{1 + x^{-3} } ^2 + \sqrt[3]{1 + x^{-3} } + 1) \\ &- \frac{1}{\sqrt 3} \arctan ( \frac{2\sqrt[3]{1 + x^{-3} } + 1}{ \sqrt 3} ) + cst \\
\end{align}

$$ \int\frac {(1+\frac1 {x^3})^{1/3}}x dx = -\sqrt[3]{1 + x^{-3} } +
 \frac{1}{3} \ln( \sqrt[3]{1 + x^{-3} } -1 )  \\ - \frac{1}{6}
 \ln(\sqrt[3]{1 + x^{-3} } ^2 + \sqrt[3]{1 + x^{-3} } + 1) -
 \frac{1}{\sqrt 3} \arctan ( \frac{2\sqrt[3]{1 + x^{-3} } + 1}{ \sqrt
 3} ) + cst $$

