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

Here is a secure and faster method when the fraction has a pole of comparatively high order:

1. 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 $$\frac 1{(x+1)^4(x^2+1)}=\frac1{u^4}\cdot\frac 1{2-2u+u^2}.$$
2. Perform the division of $1$ by $2-2u+u^2$ along increasing powers of $u$, up to order $4$: $$\begin{array}{r} \phantom{\frac12}\\ \phantom{u}\\ 2-2u+u^2\Big( \end{array}\begin{array}[t]{&&rr@{}rrrrr} \frac12&{}+\frac 12 u&{}+\frac 14u^2 \\ %\hline 1 \\ -1&{}+u&{}-\frac12u^2 \\\hline &u&{}-\frac12u^2 \\ &-u& +u^2 &{}-\frac12u^3\\ \hline &&&\frac12u^2&{}-\frac12u^3 \\ &&&-\frac12u^2&{}+\frac12u^3&-\frac14u^4 \\ \hline &&&&&-\frac14u^4 \end{array} $$
3. 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 $x$ : $$\frac 1{(x+1)^4(x^2+1)}=\frac1{2(x+1)^4}+\frac 1{2(x+1)^3} +\frac 1{4(x+1)^2}-\frac1{4(x^2+1)}.$$

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?

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.