How can one integrate $\int\frac{1}{(x+1)^4(x^2+1)} dx$? 
How can one integrate $\displaystyle\int\frac{1}{(x+1)^4(x^2+1)}\  dx$?

Attempt:
I tried partial fraction decomposition (PFD) and got lost. The method of u-substitution didn't work for me either. 
What else can I do? Can one calculate the integral without PFD?
 A: Here is a secure and faster method when the fraction has a  pole  of comparatively high order:


*

*If the pole is not $0$, as is the case here, perform the substitution $u=x+1$ and express the other factors in function  of $u$. We have to take care of $x^2+1$. The method of successive divisions yields $x^2+1=u^2-2u+2$, so we have
$$\frac 1{(x+1)^4(x^2+1)}=\frac1{u^4}\cdot\frac 1{2-2u+u^2}.$$

*Perform the division of $1$ by $2-2u+u^2$ along increasing powers of $u$, up to order $4$:
$$\begin{array}{r}
\phantom{\frac12}\\
\phantom{u}\\
2-2u+u^2\Big(
\end{array}\begin{array}[t]{&&rr@{}rrrrr}
 \frac12&{}+\frac 12 u&{}+\frac 14u^2 \\
%\hline
1 \\
-1&{}+u&{}-\frac12u^2 \\\hline
&u&{}-\frac12u^2 \\
&-u& +u^2 &{}-\frac12u^3\\
\hline
&&&\frac12u^2&{}-\frac12u^3 \\
&&&-\frac12u^2&{}+\frac12u^3&-\frac14u^4  \\
\hline
&&&&&-\frac14u^4  
\end{array} $$

*This yields the equality:
$$1=(2-2u+u^2)\bigl(\tfrac12+\tfrac 12 u+\tfrac 14u^2\bigr)-\tfrac14u^4,$$
whence the partial fractions decomposition:


$$\frac 1{u^4(2-2u+u^2)}=\frac1{2u^4}+\frac 1{2u^3} u+\frac 1{4u^2}-\frac1{4(2-2u+u^2)},$$
or with $x$ :
$$\frac 1{(x+1)^4(x^2+1)}=\frac1{2(x+1)^4}+\frac 1{2(x+1)^3} +\frac 1{4(x+1)^2}-\frac1{4(x^2+1)}.$$
A: We use a variant of the Heaviside method. Shift by one and consider
$$\frac{1}{z^4(z^2-2z+2)}\text{.}$$
Develop $1/(z^2-2z+2)$ in series about $z=0$, keeping the remainder exactly as you go:
$$\frac{1}{z^2-2z+2}=\frac{1}{2}+\frac{z}{2}+\frac{z^2}{4}-\frac{z^4}{4(z^2-2z+2)}\text{.}$$
Then
$$\frac{1}{z^4(z^2-2z+2)}=\frac{2 + 2z + z^2}{4z^4}-\frac{1}{4(z^2-2z+2)}\text{.}$$
Can you take it from here?
A: This can actually be done with very elementary math.
Step 1: Perform $u$-sub $x+1=t$, 
step 2: Perform u-sub $t=\frac{1}{z}$. The integral becomes $\int\frac{-z^4}{2z^2-2z+1} dz$ upon which long division can be performed. 
You will have to integrate couple of polynomial terms, you will also  get a natural log and a basic arctangent (after completing the square on $2z^2-2z+1$). Then you need to backsub. A bit of annoying algebra, but very elementary in terms of calculus, and no partial fraction decomposition.
