<s>[INST] Evaluating $\int\frac{x^4}{x-1} \, dx$ This question involves long division. 
I calculated the value. However, I want to ask two concept questions:
1) Why am I doing long division rather than writing out the form of the partial fraction decomposition of the function. 
2) When utilizing long division do I place a constant at the end and if not why? 
Here is my set up $\dfrac{x^4+0x^3+0x^2+0x+1}{x-1}$
 [/INST] 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:
$$\int{\frac{(y+1)^2}{y}dy}=\int\frac{y^4+4y^3+6y^2+4y+1}{y}dy$$
$$=\int y^3+4y^2+6y+4+\frac{1}{y}dy$$
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