<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] $\frac{1}{(x+1)(x+2)^2(x+3)^3} = \frac {A}{x+1} + \frac {B}{x+2} + \frac {C}{(x+2)^2} + \frac {D}{x+3} + \frac {E}{(x+3)^2} + \frac F{(x+3)^3} $
Here is a little trick.
Multiply trough by $(x+1)$
$\frac{(x+1)}{(x+1)(x+2)^2(x+3)^3} = A + \frac {B}{x+2}(x+1) + \frac {C}{(x+2)^2}(x+1) + \frac {D}{x+3}(x+1) + \frac {E}{(x+3)^2}(x+1) + \frac F{(x+3)^3}(x+1) $
And take the limit as x approaches $-1$
$\lim_\limits{x\to -1} \frac{(x+1)}{(x+1)(x+2)^2(x+3)^3} = \frac {1}{(1^2)(2^3)} = \frac 18 = A$
We can do something similar to quickly find $C, F$
$\lim_\limits{x\to -2} \frac{(x+2)^2}{(x+1)(x+2)^2(x+3)^3} =  C$
$\lim_\limits{x\to -3} \frac{(x+3)^3}{(x+1)(x+2)^2(x+3)^3} = F$
That leaves B, D, E
Multiplying through by $(x+2)^2$  and simplifying the LHS
$\frac{1}{(x+1)(x+3)^3} = \frac {A}{x+1}(x+2)^2 + B(x+2) + C + \frac {D}{x+3}(x+2)^2 + \frac {E}{(x+3)^2}(x+2)^2 + \frac F{(x+3)^3}(x+2)^2$
If we take the derivative of both sides and take the limits of as x approaches $- 2$
$\lim_\limits{x\to -2}\frac {d}{dx}\frac{1}{(x+1)(x+3)^3} =  B\\
\lim_\limits{x\to -2}\frac{- 3(x+1) - (x+3)^3}{(x+1)^2(x+3)^4} =  B\\
2 =  B$
$\lim_\limits{x\to -3}\frac {d}{dx}\frac{1}{(x+1)(x+2)^2} =  E$
And we take a second derivative to find $D$
$\lim_\limits{x\to -3}\frac {d^2}{dx^2}\frac{1}{(x+1)(x+2)^2} =  2D$
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