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
 A: Hint Substitute $u=1-x$ ......
A: HINT:
$$x^4=\{1-(1-x)\}^4=1-\binom41(1-x)+\binom42(1-x)^2-\binom43(1-x)^3+(1-x)^4$$
A: 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?
A: HINT:
$$\int\frac{x^4}{(1-x)^3}\space\text{d}x=$$
$$-\int\frac{x^4}{(x-1)^3}\space\text{d}x=$$

Do long division:

$$-\int\left[x+3+\frac{6}{x-1}+\frac{4}{(x-1)^2}+\frac{1}{(x-1)^3}\right]\space\text{d}x=$$
$$-\left[\int x\space\text{d}x+\int3\space\text{d}x+\int\frac{6}{x-1}\space\text{d}x+\int\frac{4}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\int x\space\text{d}x+3\int1\space\text{d}x+6\int\frac{1}{x-1}\space\text{d}x+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\frac{x^2}{2}+3x+6\int\frac{1}{x-1}\space\text{d}x+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$

For the integrand $\frac{1}{x-1}$, substitute $u=x-1$ and $\text{d}u=\text{d}x$:

$$-\left[\frac{x^2}{2}+3x+6\int\frac{1}{u}\space\text{d}u+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\frac{x^2}{2}+3x+6\ln\left|u\right|+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]=$$
$$-\left[\frac{x^2}{2}+3x+6\ln\left|x-1\right|+4\int\frac{1}{(x-1)^2}\space\text{d}x+\int\frac{1}{(x-1)^3}\space\text{d}x\right]$$
