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

calculusindefinite-integralsintegrationpartial fractions

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?

Best Answer

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 $x$ : $$\frac 1{(x+1)^4(x^2+1)}=\frac1{2(x+1)^4}+\frac 1{2(x+1)^3} +\frac 1{4(x+1)^2}-\frac1{4(x^2+1)}.$$

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