<s>[INST] How to integrate $\frac{1}{(x+1)(x+2)^2(x+3)^3}$? I tried to solve it with partial fraction decomposition but the expression becomes way too difficult to solve. I could only solve three of six(A-F) expressions of the partial fraction expansion.
 [/INST] In order to determine the partial fraction decomposition of $$f(x) = \frac{1}{(x+1)(x+2)^2(x+3)^3} = \frac{A_1}{x+1} + \frac{B_1}{x+2} + \frac{B_2}{(x+2)^2} + \frac{C_1}{x+3} + \frac{C_2}{(x+3)^2} + \frac{C_3}{(x+3)^3}$$ we can proceed as follows:
$$A_1 = \lim\limits_{x\to -1} (x+1)f(x) = \lim\limits_{x\to -1} \frac{1}{(x+2)^2(x+3)^3} = \frac{1}{8}$$
$$B_2 = \lim\limits_{x\to -2} (x+2)^2f(x) = \lim\limits_{x\to -2} \frac{1}{(x+1)(x+3)^3} = -1$$
$$C_3 = \lim\limits_{x\to -3} (x+3)^3f(x) = \lim\limits_{x\to -3} \frac{1}{(x+1)(x+2)^2} = -\frac{1}{2}$$
$$B_1 = \lim\limits_{x\to -2} (x+2)\left[f(x) - \frac{B_2}{(x+2)^2}\right] = \lim\limits_{x\to -2} \frac{1}{x+2}\left[(x+2)^2f(x)-B_2\right] $$ $$ = \lim\limits_{x\to -2} \frac{1}{x+2}\left[\frac{1}{(x+1)(x+3)^3}+1\right] = -\lim\limits_{x\to -2}\frac{1+(x+1)(x+3)^3}{x+2} $$ $$ = -\lim\limits_{x\to -2} \left[(x+3)^3 + 3(x+1)(x+3)^2\right] = 2 \text{ (by l'Hopital's rule)}$$
$$C_2 = \lim\limits_{x\to -3} (x+3)^2\left[f(x) - \frac{C_3}{(x+3)^3}\right] = \lim\limits_{x\to -3} \frac{1}{x+3}\left[(x+3)^3f(x) - C_3\right]$$ $$ = \lim\limits_{x\to -3}\frac{1}{x+3}\left[\frac{1}{(x+1)(x+2)^2} + \frac{1}{2}\right] = -\frac{1}{2}\lim\limits_{x\to -3}\frac{\frac{1}{2}(x+1)(x+2)^2+1}{x+3}$$ $$ = -\frac{1}{2}\lim\limits_{x\to -3}\left[\frac{1}{2}(x+2)^2 + (x+1)(x+2)\right] = -\frac{5}{4} \text{ (by l'Hopital's rule)}$$
$$C_1 = \lim\limits_{x\to -3} (x+3)\left[f(x) - \frac{C_2}{(x+3)^2} - \frac{C_1}{(x+3)^3}\right] = \frac{1}{(x+3)^2}\left[(x+3)^3f(x) - C_2(x+3) - C_3\right]$$ $$ = \lim\limits_{x\to -3} \frac{1}{(x+3)^2}\left[\frac{1}{(x+1)(x+2)^2}+\frac{5}{4}(x+3)+\frac{1}{2}\right] $$ $$ = -\frac{1}{2}\lim\limits_{x\to -3}\frac{\frac{5}{4}(x+1)(x+2)^2(x+3)+\frac{1}{2}(x+1)(x+2)^2+1}{(x+3)^2} $$ $$ = -\frac{1}{4}\lim\limits_{x\to -3} \frac{\mathrm{d}^2}{\mathrm{d}x^2}\left[\frac{5}{4}(x+1)(x+2)^2(x+3)+\frac{1}{2}(x+1)(x+2)^2+1\right] \text{ (by l'Hopital's rule twice)}$$ $$ = -\frac{17}{8}$$
The last one is more annoying to calculate unfortunately.
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