I know how to use this algorithm when I am integrating rational functions, but my textbook has omitted the actual proof for why it works. If someone could please help me with this question:
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$\begingroup$ Multiply the equation by $(x-1)^2$ then solve for $c_{1,2}$. After clearing denominators, partial fraction decomposition is reduced to linear algebra. $\endgroup$– anonFeb 22, 2013 at 0:53
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$\frac{c_1}{x-1}+\frac{c_2}{(x-1)^2}=\frac{c_1(x-1)+c_2}{(x-1)^2}$. Equating coefficients in $c_1x+(c_2-c_1)=c_1(x-1)+c_2=ax+b$ gives $c_1=a$ and $c_2-c_1=b$ so that $c_2=a+b$.
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$\begingroup$ Would this be considered a proof, I never know if something is considered "rigorous" or not? $\endgroup$– DickFeb 22, 2013 at 1:21