<s>[INST] integrate $\int \frac{x^4}{(1-x)^3}dx$ 
$$\int \frac{x^4}{(1-x)^3}$$

Because the degree of the numerator is bigger than the denominator I need to do a long division, should I open $(1-x)^3$ or is there a shorter way?
I got to $-\int x-\int 3-\int \frac{6x^2-10x-3}{(x-1)^3}$
So $6x^2-10x-3=A(x-1)^2+B(x-1)+C$
So $A=6$ And $B=2$ AND $C=-9$ which is wrong
 [/INST] First, substitute $u = 1 -x$ into the integral:
$$u = 1 - x$$
$$x = 1 - u$$
$$du = -dx$$
$$\int \frac{x^4}{(1-x)^3}dx$$
$$\int \frac{(1-u)^4}{u^3}(-du)$$
$$-\int \frac{(1-u)^4}{u^3}du$$
$$-\int \frac{(u-1)^4}{u^3}du$$
$$-\int \frac{u^4 - 4u^3 + 6u^2 - 4u + 1}{u^3}du$$
$$-\int \left(u - 4 + \frac{6}{u} - \frac{4}{u^2} + \frac{1}{u^3}\right)du$$
Can you take it from here?
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