Evaluating $\int\frac{x^4}{x-1} \, dx$

Partial fractions will only work if the degree of the numerator is strictly less than the degree of the denominator. Hence why you need to do long division. You should not add the +1 at the end on the numerator. This changes the value of the fraction.

If you remember differences of squares you can do the division without long division as:

$$\frac{x^4}{x-1}=\frac{x^4-1}{x-1}+\frac{1}{x-1}$$ $$=\frac{(x^2-1)(x^2+1)}{x-1}+\frac{1}{x-1}$$ $$=\frac{(x-1)(x+1)(x^2+1)}{x-1}+\frac{1}{x-1}$$ $$=(x+1)(x^2+1)+\frac{1}{x-1}$$

Alternatively you could substitution to make the division easier. Let $y=x-1$, hence $dy=dx$ and the integral becomes:

$$\int{\frac{(y+1)^2}{y}dy}=\int\frac{y^4+4y^3+6y^2+4y+1}{y}dy$$ $$=\int y^3+4y^2+6y+4+\frac{1}{y}dy$$

In integrating a rational function, long division or something accomplishing the same purpose is the first step. When you divide the numerator $x^4$ by $x-1$ you get a quotient of $x^3+x^2+x+1$ and a remainder of $1$; therefore $$ \frac{x^4}{x-1} = x^3 + x^2 + x + 1 + \frac 1 {x-1}. $$

This is just like what you do in arithmetic with integers. For example, when you divide $215$ by $19$ you get a quotient of $11$ and a remainder of $6$; therefore $$ \frac{215}{19} = 11 + \frac 6 {19}. $$

Where partial fractions enter is as follows: $$ \frac{x^4}{x-1} = x^3 + x^2 + x + 1 + \underbrace{\qquad\frac 1 {x-1}\qquad}_{\begin{smallmatrix} \text{Use partial fractions} \\ \text{for this part if necessary.} \end{smallmatrix}} $$

If the denominator is a first-degree polynomial, then partial fractions will not be needed. If it is a higher-degree polynomial, then the first question is how to factor it. If it is an irreducible second-degree polynomial, you'll complete the square and end up with an arctangent. If it is a reducible second-degree polynomial, then you go straight to the simplest case of partial fractions. A higher-than-second-degree polynomial can always be factored into first- and second-degree factors.