Partial Fraction Decomposition With Quadratic Term

[Cosmophile]

Hey, all! I'm learning partial fraction decomposition from Serge Lang's "A First Course in Calculus." In it, he gives the following example:

\int\frac{x+1}{(x-1)^2(x-2)}dx

He then decomposes this into the following sum:

\frac{x+1}{(x-1)^2(x-2)} = \frac{c_1}{x-1}+\frac{c_2}{(x-1)^2}+\frac{c_3}{x-2}

My question is this: On the right hand side (RHS), $x-1$ and $(x-1)^2$ appear. Why is this the case, when the original denominator only had the $(x-1)^2$? I hope this makes sense, and any help here is greatly appreciated!

 

[mathman]

If you left out the x-1 denominator term, then the numerator for (x-1)^2 would be a+bx. The expression you are given is equivalent and is easier to integrate.

 

[Cosmophile]

I'm afraid I don't really understand. Could you explain more explicitly, or direct me to a good resource on this?

 

[pwsnafu]

I've answer this in a previous thread, so read that first and then you can ask more questions here.