How to solve integral of polynomial for example $$\int {(x^2 - 3x)}^5$$
Can we solve this without expansion $$\int {(2x - 6)}^3 dx $$
the second one we can do using substitution
$u = 2x-6$
$du = 2 ~dx$
$\displaystyle \int (2x-6)^3~dx \Rightarrow \int u^3~\dfrac{du}{2} = \dfrac{u^4}{8} = \dfrac{(2x-6)^4}{8} = 2(x-3)^4$
One way or another you're going to have to expand the first one.
For the second one, I would start by factoring out $2^3$: $8 \int (x- 3)^3 dx$. Let u= x- 3 so du= dx. The integral becomes $8 \int u^3 du= \frac{8}{4} u^4+ C= 2 (x- 3)^4+ C$.
You say "the expansion of $(2x- 6)^4/8$ is not like the answer."
The expansion is $2(x^4- 12x^3+ 54x^2- 108x+ 81)= 2x^4- 24x^3+ 108x^3- 216x+ 162$. What is "the answer"?
None answered the first one. If you know that: $(x^3-3x)^5=x^{15}-15x^{13}+90x^{11}-270x^9+405x^7-243x^5$
wouldn't it be easy to find $\int {{{({x^3} - 3x)}^5}}~?$ See here.