Evaluation of $\int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$ Evaluation of $\displaystyle \int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$
$\bf{My\; Try::}$ Let $x=t^4\;,$ Then $dx = 4t^3dt$
So Integral is $\displaystyle \int\frac{\sqrt[3]{t^4+t}}{t^2} \cdot 4t^3dt$
So Integral is $\displaystyle 4\int t^{\frac{7}{3}}\cdot (1+t^{-3})^{\frac{1}{3}}$
Now How can i solve after that
Help me
Thanks

Let $\mathcal{I}$ be the integral. You can actually evaluate it using the substitution $x = t^4$.
$$\mathcal{I} = 
\int\frac{\sqrt[3]{x + \sqrt[4]{x}}}{\sqrt{x}}dx
= \int\frac{\sqrt[3]{(t^3 + 1)t}}{t^2}4t^3dt
= 4\int\sqrt[3]{1+t^3}t^{4/3}dt
$$
For $|x| < 1$, we can expand the integrand at RHS using following expansion
$$\frac{1}{(1-t)^\gamma} = \sum_{k=0}^\infty \frac{(\gamma)_k}{k!} t^k$$
where $(\gamma)_k = \gamma(\gamma+1)\cdots(\gamma+k-1)$ is the rising Pochhammer symbol. This gives us
$$
\mathcal{I} 
= 4\int\left(\sum_{k=0}^\infty \frac{(-1)^k (-\frac13)_k}{k!}t^{3k}\right)t^{4/3}dt
= 4 \sum_{k=0}^\infty \frac{(-1)^k (-\frac13)_k}{k!}\frac{t^{3k+7/3}}{3k+7/3}
$$
Using another identity
$$\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma}{\gamma+k}$$
We can transform above expression to
$$\mathcal{I} 
= \frac{12}{7} t^{\frac73}
\sum_{k=0}^\infty \frac{(-\frac13)_k}{k!}\frac{(\frac79)_k}{(\frac{16}{9})_k}(-t^3)^k
$$
The expansion in the right is that for a 
hypergeometric function ${}_2F_1$. As a result, up to an integration constant, we have
$$\mathcal{I}
= \frac{12}{7} t^{\frac73} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -t^3 \right)
= \frac{12}{7} x^{\frac{7}{12}} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -x^{\frac34} \right)$$
