Integral of $\frac{{x^{1/2}}+3}{2+{x^{1/3}}}$ I want to solve this integral and think about doing the following steps:
$1)\quad t=x^{1/3}$
$2)\quad x=t^3$
$3)\quad dx=2t^3\,dt$
How I can show $\sqrt{x}$ as $t$?
$$\frac{x^{1/2}+3}{2+x^{1/3}}$$
Thanks!
 A: The substitution $t=x^{\frac{1}{6}}$ is better. Then
$$\sqrt{x}=t^3,\;\;\sqrt[3]{x}=t^2 \\
dx=6t^5 dt,$$
so
$$
\int{\frac{\sqrt{x}+3}{2+{\sqrt[3]{x}}}\ dx}=6\int{\frac{(t^3+3)t^5}{2+t^2}\ dt}.
$$
A: If you want to use your substitution $t=x^{1/3}$:
$$
\begin{equation*}
I=\int \frac{x^{1/2}+3}{2+x^{1/3}}dx=3\int \frac{t^{2}(
t^{3/2}+3) }{2+t}\,dt,\tag{1}
\end{equation*}
$$
then you could use the additional substitution $u=t^{1/2}=x^{1/6}$ to obtain
$$
\begin{equation*}
I=6\int \frac{u^{8}+3u^{5}}{u^{2}+2}\,du.\tag{2}
\end{equation*}
$$
To evaluate this integral rewrite the integrand as
$$\frac{u^{8}+3u^{5}}{u^{2}+2}=u^{6}-2u^{4}+3u^{3}+4u^{2}-6u-8+\frac{12u+16}{u^{2}+2},$$
using polynomial long division.
However, in general when the integrand is of the form $$f(x)=g(x,x^{p/q},x^{r/s},\ldots ),$$ with $p,q,r,s,\ldots\in \mathbb{N}$, the standard substitution is $x=t^{k}$, where $k=\operatorname{lcm}(q,s,\ldots )$. In the present case as pointed out by M. Strochyk the direct substitution is thus  $x=t^{6}$, which yields the same integral as $(2)$:
$$
\begin{equation*}
I=\int \frac{x^{1/2}+3}{2+x^{1/3}}dx=6\int \frac{t^{8}+3t^{5}}{t^{2}+2}
\,dt,\qquad x=t^{6}\tag{3}.
\end{equation*}
$$
