<s>[INST] 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?
 [/INST] 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 $x$ :
$$\frac 1{(x+1)^4(x^2+1)}=\frac1{2(x+1)^4}+\frac 1{2(x+1)^3} +\frac 1{4(x+1)^2}-\frac1{4(x^2+1)}.$$
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