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.
Substitute $\text{u}=x-1$:
$$\int\frac{x^4}{x-1}\space\text{d}x=\int\frac{\left(1+\text{u}\right)^4}{\text{u}}\space\text{d}\text{u}=\int\left(\text{u}^3+4\text{u}^2+6\text{u}+\frac{1}{\text{u}}+4\right)\space\text{d}\text{u}=$$ $$\int\text{u}^3\space\text{d}\text{u}+4\int\text{u}^2\space\text{d}\text{u}+6\int\text{u}\space\text{d}\text{u}+\int\frac{1}{\text{u}}\space\text{d}\text{u}+4\int1\space\text{d}\text{u}=$$ $$\frac{\text{u}^4}{4}+4\cdot\frac{\text{u}^3}{3}+6\cdot\frac{\text{u}^2}{2}+\ln\left|\text{u}\right|+4\text{u}+\text{C}=$$ $$\frac{\left(x-1\right)^4}{4}+\frac{4\left(x-1\right)^3}{3}+3\left(x-1\right)^2+\ln\left|x-1\right|+4\left(x-1\right)+\text{C}$$
$$\int\frac{x^4-1+1}{x-1}dx=\int\frac{x^4-1}{x-1}dx+\int\frac{1}{x-1}dx$$ note $$x^4-1=(x-1)(x^3+x^2+x+1)$$