Is it necessary to write limits for a substituted integral? To solve the following integral, one can use u-substitution:
$$\int_2^3 \frac{9}{\sqrt[4]{x-2}} \,dx,$$
With $u = \sqrt[4]{x-2}$, our bounds become 0 and 1 respectively. Thus, we end up with the following:
$$36\int_0^1{u^2} \,du$$
In the first case, the lower bound is a vertical asymptote so we would have to use limits to find the answer. However, in the second case there's no longer an asymptotal bound - would you still have to write the limits since the original function would've needed limits, or can you just solve this by plugging in the substituted bounds? I know the final answer will be the same either way, but I want to know if it can be considered correct to exclude the limits in the 2nd integral from a technical perspective since the function has been changed. Many thanks in advance!

Suppose you must.  Then we have $$9\lim_{a\to2^+}\int_a^3\frac1{\sqrt[4]{x-2}}\,dx.$$
Let $u=x-2$.  Then we have $$9\lim_{a\to2^+}\int_{a-2}^1 u^{-1/4}\, du.$$
By the power rule, we have $$9\lim_{a\to2^+} \left.\frac43u^{3/4}\right]_{a-2}^1=9\left[\frac43(1-\lim_{a\to2^+}(a-2)^{3/4})\right]=9\left[\frac43(1-0)\right]=12.$$
Notice that it does not make a difference whether you use $\lim_{a\to2^+}(a-2)$ or $0$.
