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`:
(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:
(Notice I'm not mixing up `x`'s and `u`'s in any one expression here.)
Now we integrate with respect to `u`:
Finally, we express everything in terms of `x`, since that's the variable we started with: