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.
 A: $\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 $D$
$\lim_\limits{x\to -3}\frac {d^2}{dx^2}\frac{1}{(x+1)(x+2)^2} =  2D$
A: Following @paulinho's suggestion, we want to write $\frac{1}{u^2(u-1)(u+1)^3}$ as a sum of partial fractions. The rest we can build by repeatedly using how to write the reciprocal of a quadratic with partial fractions. Note that $$\frac{1}{(u-1)(u+1)}=\frac12\left(\frac{1}{u-1}-\frac{1}{u+1}\right)\\\implies\frac{1}{(u-1)(u+1)^2}=\frac14\left(\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{(u+1)^2}\right)\\\implies\frac{1}{(u-1)(u+1)^3}=\frac18\left(\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{(u+1)^2}-\frac{4}{(u+1)^3}\right)\\\implies\frac{1}{u^2(u-1)(u+1)^3}=\frac18\left(\frac{1}{u^2(u-1)}-\frac{1}{u^2(u+1)}-\frac{2}{u^2(u+1)^2}-\frac{4}{u^2(u+1)^3}\right).$$You can do the rest yourself with such observations as$$\frac{1}{u^2(u\pm 1)}=\pm\frac{1}{u}\left(\frac{1}{u}-\frac{1}{u\pm 1}\right)=\pm\frac{1}{u^2}\mp\frac{1}{u}\pm\frac{1}{u\pm 1},\\\frac{1}{u^2(u+1)^2}=\frac{1}{u^2}-\frac{2}{u}+\frac{2}{u+1}+\frac{1}{(u+1)^2}.$$
A: The substitution
$$x=-\left( 3+\frac{2}{u} \right)$$
reduces the integral to
$$\begin{align}
  & =\frac{1}{8}\int{\frac{{{u}^{4}}}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du} \\ 
 & =\frac{1}{8}\int{u-5+\frac{17{{u}^{2}}+36u+20}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du} \\ 
 & =\frac{1}{8}\int{u-5+\frac{1}{u+1}+\frac{16}{u+2}-\frac{16}{{{\left( u+2 \right)}^{2}}}du} \\ 
\end{align}$$
and you can see that partial fraction decomposition becomes much easier. 
A: When partial fraction decomposition becomes a bit overwhelming, you can apply the Horowitz-Ostrogradsky algorithm ! [Manuel Bronstein - Symbolic Integration I]
It is very mechanical, only the calculation of $H$ is tedious, but the rest is quite easy.
So we start with $$\frac AD=\frac{1}{(x+1)(x+2)^2(x+3)^3}$$
$A=1$
$D=(x+1)(x+2)^2(x+3)^3$
$D\,'=(x+2)(x+3)^2(6x^2+22x+18)$
$D^-=\gcd(D,D\,')=(x+2)(x+3)^2$
$D^*=D/D^-=(x+1)(x+2)(x+3)$
$B=\sum_{i=0}^{\deg(D^-)-1}b_ix^i=b_0+b_1x+b_2x^2$
$C=\sum_{i=0}^{\deg(D^*)-1}c_ix^i=c_0+c_1x+c_2x^2$

And let's identify to the null polynomial $$\forall x:\quad H(x)=A-B\,'D^*+BD^*{D^{-}}'/D^--CD^-=0$$

$H(x)=-c_2x^5+(b_2-c_1-8c_2)x^4+(2b_1-2b_2-c_0-8c_1-21c_2)x^3+(3b_0+4b_1-15b_2-8c_0-21c_1-18c_2)x^2+(10b_0-4b_1-12b_2-21c_0-18c_1)x+(7b_0-6b_1-18c_0+1)$

This system solves to $\begin{cases}B=\frac 94x^2+\frac{25}{2}x+17\\C=\frac 94x+\frac 52\end{cases}$
And the formula says :
$$\int \frac AD\mathop{dx}=\frac{B}{D^-}+\int\frac{C}{D^*}\mathop{dx}=\frac{9x^2+50x+68}{4(x+2)(x+3)^2}+\int\frac{9x+10}{4(x+1)(x+2)(x+3)}\mathop{dx}$$

The last part is still solved by partial fraction decomposition, but is much simpler:
$\int=-\frac{17}8\ln(x+3)+2\ln(x+2)+\frac 18\ln(x+1)$
A: 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.
