<s>[INST] Find the integral of $\int \frac{x^4}{x-1}$ I was not happy with the responses so I went ahead and solved the integral to get feedback on the answer:
Dividing as the book shows us:
we get $x^4$ divided by $x-1$ which gives us the quotient plus the remainder over the divisor which is: $x^3 + \frac{x^3}{x-1}$
$\int x^3+\frac{x^3}{x-1} \to \frac{1}{4}x+x^3ln(\vert x-1 \vert)$
Instead of using shortcuts I want feedback on how it should be done with the process in the book.
 [/INST] $$\int\frac{x^4-1+1}{x-1}dx=\int\frac{x^4-1}{x-1}dx+\int\frac{1}{x-1}dx$$
note
$$x^4-1=(x-1)(x^3+x^2+x+1)$$
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