<s>[INST] How to find the integral with $\sqrt [ 3 ]{ x } +\sqrt [ 4 ]{ x } $ in the denominator? How to evaluate $$\int { \frac { 1 }{ \sqrt [ 3 ]{ x } +\sqrt [ 4 ]{ x }  }  } +\frac { \log { (1+\sqrt [ 6 ]{ x } ) }  }{ \sqrt [ 3 ]{ x } +\sqrt { x } } dx$$ I'm not being able to make the right substitution.Help please!
 [/INST] when you have so many fractional powers, to simplify things try to go for a substitution that can clear up all the powers (usually the LCM of all denominators of various powers will do the job). For example, here let $x=t^{12}$, then you get
\begin{align*}
\int {\frac{1}{\sqrt[3]{x} +\sqrt[4]{x}}} +\frac{\log{(1+\sqrt[6]{x})}}{\sqrt[3]{x}+\sqrt{x}}\, dx & = \int\left[ \frac{1}{t^4+t^3}+ \frac{\log(1+t^2)}{t^4+t^6}\right] \, (12t^{11}) \, dt\\
&= 12 \left[\int \frac{t^8}{1+t} \, dt + \int  \frac{t^7 \log(1+t^2)}{1+t^2} \, dt\right]\\
& = 12 \left[\int \frac{t^8-1+1}{1+t} \, dt + \int  \frac{t^7 \log(1+t^2)}{1+t^2} \, dt\right]\\
& = 12 \left[\int \frac{1}{1+t} \, dt + \int \frac{t^8-1}{t+1} \, dt + \int  \frac{t^7 \log(1+t^2)}{1+t^2} \, dt\right]
\end{align*}
Hopefully now you can solve this. First integral is easy, second is simple factorization, the third is substitution and using integration by parts.
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