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?

set $$x=1-\frac{1}{u+1}$$ to get
$$\int{\frac{{{x}^{6}}}{{{(x-2)}^{2}}{{(1-x)}^{5}}}dx}=\int{\frac{{{u}^{6}}}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du}$$
long division for the integrand gives:
$${{u}^{3}}-5{{u}^{2}}+17u-49+\frac{129{{u}^{2}}+324u+196}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}$$ 
