The following text is from Mathematics for Class XII by Dr. R.D.Sharma, chapter "Indefinite Integrals", topic "Integration of Rational Algebraic Functions by using Partial Fractions", sub topic "Case II: When the denominator $g(x)$ [in $\frac{f(x)}{g(x)}$] is expressible as the product of the linear factors such that some of them are repeating.":
Let $g(x)=(x-a)^k(x-a_1)(x-a_2)\dots(x-a_r).$ Then we assume that
$$\frac{f(x)}{g(x)}=\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\frac{A_3}{(x-a)^3}+\dots+\frac{A_k}{(x-a)^k}+\frac{B_1}{x-a_1}+\frac{B_2}{x-a_2}+\dots+\frac{B_r}{x-a_r}$$
i.e., corresponding to non-repeating factors we assume as in Case I [When denominator is expressible as the product of non-repeating linear factors] and for each repeating factor $(x-a)^k$, we assume partial fractions
$$\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\frac{A_3}{(x-a)^3}+\dots+\frac{A_k}{(x-a)^k}$$
where $A_1,A_2,\dots,A_k$ are constants.
What is the reasoning behind this particular choice of partial fractions in this case? Why do we want to consider all positive integral powers of $(x-a)$ upto $k$? What if we assume the partial fraction with only the first term $\frac{A_1}{x-a}$ like Case I? Why not consider only the first and last terms in the given partial fraction expression? I think it'll save a lot of time. Will any essential data will be lost as we neglect the rest of the terms in the expression?
Best Answer
Short answer: you need the higher order terms because without them you do not have enough flexibility to write an expansion you can integrate.
For example, how you would you express $$ \frac{1}{x^2 (x-1)} $$ in the form $$ \frac{A}{x} + \frac{B}{x-1} ? $$
When you clear fractions and try to solve for the constants the polynomial degrees don't match.
If you tried the form $$ \frac{A}{x^2} + \frac{B}{x-1} $$ you would have the right denominator but not enough freedom to match the coefficients in the numerator.