We use, as a starting point, the substitution

`u = 2x^4- 5`.

Why? Because `2x^4- 5` is the expression in brackets in the question.

Now differentiating `u` gives: `du = 8x^3\ dx`

Our question has only **one** `x^3\ dx` (not 8 of them) so we need
to divide both sides by `8`:

`1/8du=x^3\ dx`

(Now the right hand side is the same as what we have in the question, `x^3\ dx`.)

We can now rewrite our question as:

`int(2x^4-5)^6x^3\ dx`

`=1/8intu^6\ du`

(Notice I'm not mixing up `x`'s and `u`'s in any one expression here.)

Now we integrate with respect to `u`:

`=1/8xxu^7/7+K`

`=u^7/56+K`

Finally, we express everything in terms of `x`, since that's the variable we started with:

`=((2x^4-5)^7)/56+K`