Any shortcuts for integrating $\frac{x^6}{(x-2)^2(1-x)^5}$ by partial fractions? Is there a faster way to get the partial fraction decomposition of this $\frac{x^6}{(x-2)^2(1-x)^5}$?
$\frac{x^6}{(x-2)^2(1-x)^5} = \frac{A_1}{x-2} + \frac{A_2}{(x-2)^2} + \frac{B_1}{1-x} + \frac{B_2}{(1-x)^2} + \frac{B_3}{(1-x)^3} + \frac{B_4}{(1-x)^4} + \frac{B_5}{(1-x)^5}$
It's fairly easy to get $A_2$ and $B_5$, just by "covering up" $x-2$ and substituting $x=2$ and "covering up" $1-x$ and substituting $x=1$. So $A_2 = 64$ and $B_5=1$. But to proceed from there is there no other faster way other than to mulitply both sides of the equation by $(x-2)^2(1-x)^5$ and compare coefficients, which would result in a system of 6 equations with 6 unknowns?

The "cover-up" rule gives correct answers but it suffers greatly from lack of rigour, in fact no rigour at all. Here's a more mathematically proper way to find the coefficients.First multiply through by $$(x-2)^2(1-x)^5$$ Your new equation is
 $$x^6=A_1(x-2)(1-x)^5+A_2(1-x)^5+B_1(1-x)^4(x-2)^2+B_2(1-x)^3(x-2)^2+B_3(1-x)^2(x-2)^2+B_4(1-x)(x-2)^2+B_5(x-2)^2$$
Note that the original equation is true for all $x$ except 2 or 1 iff your new equatiion is true for all $x$ except possibly 2 or 1 iff your new equation is true for all x.[Two polynomials in one variable are equal for all $x$ iff they are equal for infinitely many $x$.There are infinitely many $x$ other than 1 or 2.]In this new equation put $x=2$ to find $A_1$ and then $x=1$ to find $B_5$Now take the derivative. $$6x^5=A_1(1-x)^5-5(1-x)^4(A_1(x-2)+A_2)+(-4B_1(1-x)^3-3B_2(1-x)^2-2B_3(1-x)-B_4)(x-2)^2+2(B_1(1-x)^4+B_2(1-x)^3+B_3(1-x)^2+B_4(1-x)+B_5)(x-2)$$ Put $x=2$ in the derived equation. Since you already know $A_2$ you can find $A_1$ Put $x=1$ in the derived equation. Since you already know $B_5$ you can find $B_4$. Another derivative will give you $B_3$ and so on.
