A change of variable for $\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx$ Question 1

I would like to understand the following change of variable

for $\phi(x)=nx,$ $\ d\phi=n\cdot dx,$ and $\chi_{E}(y)$ the usual characteristic function of a set $E$:
$$\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx= \int_0^{+\infty}\frac{\phi(x)}{n^{2/3}+n^{1/6}\phi(x)}\chi_{(0,n)}(\phi(x)) \ d\phi$$
Question 2
Is there a more slick way to evaluate the integral 
$$\lim_{n\to +\infty}\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx$$
 A: I can answer your question 1. (I don't understand your question 2, in a similar vein to @joriki's comment, and I don't understand your reply to that comment. Maybe if you elaobrate further in the comment, I will understand.)
If you make the substitution $u=nx$, then you would have
$$\int_0^1\frac{n^{4/3}x}{1+n^{5/2}x^3}dx=\int_{u=0}^{u=n}\frac{n^{1/3}u}{1+n^{-1/2}u^3}\frac{du}{n}=\int_{0}^{n}\frac{u}{n^{2/3}+n^{1/6}u^3}du$$
Now introduce that function $\chi$. Its output is $1$ if the input is in $[0,n]$, and $0$ otherwise.
$$\int_{0}^{n}\frac{u}{n^{2/3}+n^{1/6}u^3}\chi(u)\,du$$
That's fine because we just multiplied by $1$. But also we can extend the upper limit of the integral to $\infty$ because over $(n,\infty)$ this would be multiplying by $0$, and therefore not changing the value of the integral.
$$\int_{u=0}^{u=\infty}\frac{u}{n^{2/3}+n^{1/6}u^3}\chi(u)\,du$$
The posted version uses $\phi(x)$ in place of $u$, which is fine. Except I wonder if you left out an exponent of $3$ in the denominator when typing up the question.
A: As you've noted in the comments, this can be easily solved with DCT. However, to follow what you've tried to gone through with, we could substitute $u=nx$ to get $x=u/n$ and $\mathrm dx=\mathrm du/n$, which give
$$\int_0^1\frac{n^{4/3}x}{1+n^{5/2}x^3}~\mathrm dx=\int_0^n\frac{n^{4/3}n^{-1}u}{1+n^{5/2}(n^{-1}u)^3}~\frac{\mathrm du}n$$
Expanding out the denominator:
$$1+n^{5/2}(n^{-1}u)^3=1+n^{5/2}n^{-3}u^3=1+n^{-1/2}u^3$$
and the numerator:
$$n^{4/3}n^{-1}u=n^{1/3}u$$
which, put together and dividing by $n$ becomes
$$\int_0^n\frac{n^{4/3}n^{-1}u}{1+n^{5/2}(n^{-1}u)^3}~\frac{\mathrm du}n=\int_0^n\frac{u}{n^{2/3}+n^{1/6}u^3}~\mathrm du$$
or equivalently,
$$\int_0^n\frac{u}{n^{2/3}+n^{1/6}u^3}~\mathrm du=\int_0^\infty\frac{u}{n^{2/3}+n^{1/6}u^3}\chi_{(0,n)}(u)~\mathrm du$$
We can factor $n^{1/6}$ out from the denominator and bound the integral by
$$\int_0^\infty\frac{u}{n^{2/3}+n^{1/6}u^3}\chi_{(0,n)}(u)~\mathrm du\le\int_0^\infty\frac{n^{-1/6}u}{1+u^3}~\mathrm du$$
where $\displaystyle\int_0^\infty\frac{u~\mathrm du}{1+u^3}$ converges by seeing it is bounded on $[0,1]$ and comparing the rest to $\displaystyle\int_1^\infty\frac{u~\mathrm du}{u^3}$.
