Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?
 A: Lengthy but not impossible. We have for sure
$$\begin{eqnarray*} f(x)&=&\frac{1}{(x+3)(x+4)^2(x+5)^3}\\&=&\frac{A}{x+3}+\frac{B}{x+4}+\frac{C}{x+5}+\frac{D}{(x+4)^2}+\frac{E}{(x+5)^2}+\frac{F}{(x+5)^3}\end{eqnarray*}\tag{1}$$
where
$$ F=\lim_{x\to -5}(x+5)^3 f(x) = \lim_{x\to -5}\frac{1}{(x+3)(x+4)^2} = -\frac{1}{2},$$
$$ D=\lim_{x\to -4}(x+4)^2 f(x) = \lim_{x\to -4}\frac{1}{(x+3)(x+5)^3} = -1,$$
$$ A = \lim_{x\to -3}(x+3) f(x)=\lim_{x\to -3}\frac{1}{(x+4)^2 (x+5)^3} = \frac{1}{8}.\tag{2} $$
By setting $g(x)=f(x)-\frac{F}{(x+5)^3}-\frac{D}{(x+4)^2}-\frac{A}{(x+3)}$ it follows that:
$$ \begin{eqnarray*}g(x)&=&\frac{20-3x-x^2}{8(x+4)(x+5)^2}\\&=&\frac{B}{x+4}+\frac{C}{x+5}+\frac{E}{(x+5)^2}\end{eqnarray*}\tag{3}$$
where
$$ E = \lim_{x\to -5}g(x)(x+5)^2 = -\frac{5}{4}, $$
$$ B = \lim_{x\to -4}g(x)(x+4) = 2\tag{3} $$
and $C=-\frac{17}{8}$ follows from computing $g(x)-\frac{E}{(x+5)^2}-\frac{B}{x+4}$, or from a general property of the partial fraction decomposition of "rapidly decaying" meromorphic functions, that in this case grants $C=-(A+B)$, since $A,B,C$ are the residues of $f(x)$ at its poles. You may deduce the same by noticing that $\lim_{x\to +\infty} x\cdot f(x)=0$.
It follows that:
$$ \int\frac{dx}{(x+3)(x+4)^2(x+5)^3}\\=C+\frac{1}{8} \left(\frac{8}{x+4}+\frac{2}{(x+5)^2}+\frac{10}{x+5}+\log(x+3)+16\log(x+4)-17\log(x+5)\right).$$
A: Splitting into partial fractions can also be done this way:
Let $u = (x+3)(x+5)$ and $v = (x+4)^2$. Then $v-u = 1$ and hence 
\begin{align*}
\frac{1}{(x+3)(x+4)^2(x+5)} &= \frac{v-u}{uv} \\
&= \frac{1}{u} - \frac{1}{v} \\
&= \frac{1}{(x+3)(x+5)} - \frac{1}{(x+4)^2}\\
&= \frac{1}{2}\frac{1}{x+3} - \frac{1}{2}\frac{1}{x+5} - \frac{1}{(x+4)^2}
\end{align*} 
Thus
\begin{align*}
\frac{1}{(x+3)(x+4)^2(x+5)^3} & = \frac{1}{(x+5)^2}\left(\frac{1}{2}\frac{1}{x+3} - \frac{1}{2}\frac{1}{x+5} - \frac{1}{(x+4)^2}\right)\\
&= \frac{1}{2}\frac{1}{(x+3)(x+5)^2} - \frac{1}{2}\frac{1}{(x+5)^3} - \frac{1}{(x+4)^2(x+5)^2} \\
&= \frac{1}{4}\left(\frac{(x+5)-(x+3)}{(x+3)(x+5)^2}\right)- \frac{1}{2}\frac{1}{(x+5)^3} - \frac{((x+5) - (x+4))^2}{(x+4)^2(x+5)^2} \\
&= \frac{1}{4}\left(\frac{1}{(x+3)(x+5)}-\frac{1}{(x+5)^2}\right) - \frac{1}{2}\frac{1}{(x+5)^3} \\
& \qquad \qquad - \left(\frac{1}{(x+4)^2} - \frac{2}{(x+4)(x+5)} + \frac{1}{(x+5)^2}\right)\\
&=\frac{1}{4}\left(\frac{1}{2}\left(\frac{1}{(x+3)} - \frac{1}{(x+5)}\right)-\frac{1}{(x+5)^2}\right) - \frac{1}{2}\frac{1}{(x+5)^3} \\
& \qquad \qquad - \left(\frac{1}{(x+4)^2} - \frac{2}{(x+4)} + \frac{2}{(x+5)} + \frac{1}{(x+5)^2}\right)\\
&= \frac{1/8}{x+3}+\frac{2}{x+4} - \frac{1}{(x+4)^2} - \frac{17/8}{x+5} - \frac{5/4}{(x+5)^2} - \frac{1/2}{(x+5)^3}
\end{align*}
A: Lazy people like me will do this as follows. The function
$$f(x) = \frac{1}{(x+3)(x+4)^2(x+5)^3}$$
has a simple pole at $x = -3$. The principal part of the Laurent expansion around this point (i.e. the sum of the singular terms in the expansion) is:
$$s_1(x) = \frac{1}{8}\frac{1}{x+3}$$
If you calculate the principal parts of the Laurent expansion around the other singular points and add them up, then that's the partial fraction expansion, because the difference between that sum and $f(x)$ won't have any singularities left. But this is then a rational function without any singularities which is therefore a polynomial. But it will tend to zero in the limit of $x$ to infinity, therefore that polynomial must be identical to zero.
Now, we can save the work needed to actually calculate the Laurent expansion around $x = -4$ and $x = -5$. We can instead exploit the fact that $f(x)$ for large $x$ tends to zero as $x^{-6}$. This means that terms of lower order in $x^{-1}$ must cancel out in the partial fraction expansion. And instead of negative powers of $x$ we can just as well consider expanding in negative powers of $x+a$ for some suitable $a$. 
Then directly calculating the most singular terms of the principal parts amounts to a trivial calculation, see eq. (2) in Jack D'Aurizio's answer. Let's take from there that the principal part of the Laurent expansion around $x = -4$ is of the form:
$$s_2(x) = \frac{B}{x+4} -\frac{1}{(x+4)^2}$$
If we use the fact that the coefficient of $(x+5)^{-4}$ in the large $x+5$ expansion of $f(x)$ is zero, then since the principal part of the Laurent expansion around $x = -5$ doesn't yield any contributions to this term, we get an equation involving only $B$. The principal part of the Laurent expansion around $x = -4$, $s_2(x)$, yields two terms, one comes from re-expanding the term
$$-\frac{1}{(x+4)^2} = -\frac{1}{(x+5)^2}\frac{1}{\left(1-\frac{1}{x+5}\right)^2}=-\frac{1}{(x+5)^2} - \frac{2}{(x+5)^3}- \frac{3}{(x+5)^4}+\cdots$$
which therefore yields the contribution:
$$-\frac{3}{(x+5)^4}\tag{1}$$
Then we have a contribution coming from the unknown term of the same principal part:
$$\frac{B}{x+4} = \frac{B}{x+5} \frac{1}{1 - \frac{1}{x+5}} = \frac{B}{x+5} +\frac{B}{(x+5)^2} + \frac{B}{(x+5)^3}+ \frac{B}{(x+5)^4}+\cdots $$
So, we get the contribution:
$$\frac{B}{(x+5)^4}\tag{2}$$
Finally, we have a contribution coming from $s_1(x)$:
$$\frac{1}{8}\frac{1}{x+3} = \frac{1}{8}\frac{1}{x+5}\frac{1}{1-\frac{2}{x+5}} = \frac{1}{8}\frac{1}{x+5} + \frac{1}{4}\frac{1}{(x+5)^2}+ \frac{1}{2}\frac{1}{(x+5)^3}+\frac{1}{(x+5)^4}\cdots$$
This thus yields the contribution:
$$\frac{1}{(x+5)^4}\tag{3}$$
Adding up the contributions (1), (2), (3), and (3) must yield zero, therefore we have $B = 2$.
We can then similarly proceed to calculate the principal part of the Laurent expansion around $x=-5$ by considering the coefficient of $(x+5)^{-3}$, $(x+5)^{-2}$, and $(x+5)^{-1}$, but we don't need to do the first one as we can obtain this from a straightforward limit computations, see eq.(2) in Jack D'Aurizio's answer. Let's take that result from there and write the principal part as:
$$s_3(x) = \frac{C}{x+5} + \frac{E}{(x+5)^2} -\frac{1}{2}\frac{1}{(x+5)^3}$$
To calculate $E$ we can then consider the coefficient of $(x+5)^{-2}$ in the large $x+5$ expansion of $f(x)$ and equate that to zero. Then $s_3(x)$ yields the contribution:
$$\frac{E}{(x+5)^2} \tag{4}$$
And $s_2(x)$ yields the contribution:
$$\frac{1}{(x+5)^2} \tag{5}$$
And $s_1(x)$ yields:
$$\frac{1}{4}\frac{1}{(x+5)^2} \tag{6}$$
Adding up (4), (5), and (6) and equating to zero yields $E = -\frac{5}{4}$
Finally $C$ follows from equating the coefficient of $(x+5)^{-1}$ to zero. The contribution from $s_3(x)$ is:
$$\frac{C}{x+5} \tag{7}$$
The contribution from $s_2(x)$ is:
$$\frac{2}{x+5} \tag{8}$$
The contribution from $s_1(x)$ is:
$$\frac{1}{8}\frac{1}{x+5} \tag{9}$$
Adding up (7), (8), and (9) and equating the result to zero yields $C = -\frac{17}{8}$.
