Was solving some differential equations and came upon this integral:
$$\int\frac 1{x(x+1)^2} dx$$
Looked it up on wolframalpha and it can be decomposed to:
$$\frac 1x-\frac{1}{(x+1)^2}-\frac{1}{x+1}$$
How do we proceed to decompose polynomials of degree bigger than $2$?
$$\frac 1x-\color{red}{\frac{1}{(x+1)^2}}-\frac{1}{x+1}$$
Why do you want to proceed to decompose?, you can solve it easly.
$$\int\frac 1x dx-\color{red}{\int \frac{1}{(x+1)^2}dx}-\int \frac{1}{x+1}dx$$
$$=\ln|x|+\frac{1}{x+1}-\ln|x+1|+C$$
For $\color{red}{\int \frac{1}{(x+1)^2}dx}$
Set: $t=x+1$ and $dt=dx$
$$\int\frac{1}{t^2}dt\Longrightarrow =-\frac{1}{t}+C=-\frac{1}{x+1}+C$$
For fraction with this form
$$\frac{P(x)}{ (x+2)(x+3)^5}$$
the decompose should looks like:
$${A \over x+2}+{B \over x+3}+{C \over (x+3)^2}+{D \over (x+3)^3}+{E \over (x+3)^4}+{F \over (x+3)^5}$$