How To Integrate 1/sqrt(x) + x^(1/3)

 

\[\begin{array}{l} \smallint \frac{{{\rm{d}}x}}{{\sqrt x + \sqrt[{\bf{3}}]{x}}}\\\\ By\;Substitution\\ \;\;\;\;\;\;Suppose \Rightarrow {z^6} = x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;,\;note \ldots LCD\left[ {2,3} \right] = 6\\ 6{z^5}dz = dx\\ \smallint \frac{{6\;{z^5}}}{{{z^3} + {z^2}}}dz\\ 6\smallint \frac{{\;{z^3}}}{{{z^\;} + 1}}dz \end{array}\]

\[\begin{array}{l} 6\smallint {z^2} - z + 1 - \frac{{\;1}}{{{z^\;} + 1}}dz\\ \Rightarrow 6\left( {\;\frac{{{z^3}}}{3} - \frac{{{z^2}}}{2} + z - \ln \left| {z + 1} \right|\;\;} \right) + c\\ \Rightarrow 6\left( {\;\frac{{\sqrt x }}{3} - \frac{{\sqrt[3]{x}}}{2} + \sqrt[6]{x} - \ln \left| {\sqrt[6]{x} + 1} \right|\;\;} \right) + c \end{array}\]