# 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?

• I don't know if there another more or less elementary way to integrate that thing without partial fractions, but I think it is unlikely it'll be much easier than that... – DonAntonio Jul 26 '19 at 18:53
• Mathematica gives: $-\frac{3 x^2+9 x+3 (x+1)^3 \tan ^{-1}(x)+8}{12 (x+1)^3}$. – David G. Stork Jul 26 '19 at 19:01
• Well, just as said: that substitution requires from you to find out what $\;1+x\;$ and $\;1+x^2\;$ is in terms of $\;t\;$ . Not sure how faster/easier that is instead of partial fractions from the beginning. Perhaps it is just a matter of getting used to this or that. – DonAntonio Jul 26 '19 at 20:25
• @Don. That's true. My approach is just to avoid PFD, but I don't think it is quicker at all. The OP wanted a method without PFD but in his integral, I am not sure why. On the other hand, the u-sub x=1/t is very effective where a numerator is a constant and the denominator is, say x^9-x. Then PFD would be brutal. So my u-sub method is a method that one should have in his "back pocket", hence my post. – imranfat Jul 28 '19 at 16:13
• where are you getting lost with the partial fractions? – Doug M Sep 26 '19 at 1:20

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?

• How exactly does Heaviside method work and where can I read up on it? – user671231 Jul 26 '19 at 19:26
• @StackUpPhysics See here. – J.G. Jul 26 '19 at 21:10
• @J.G. Thanks for the link – user671231 Jul 26 '19 at 21:14

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.