4
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ 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... $\endgroup$ – DonAntonio Jul 26 '19 at 18:53
  • 1
    $\begingroup$ Mathematica gives: $-\frac{3 x^2+9 x+3 (x+1)^3 \tan ^{-1}(x)+8}{12 (x+1)^3}$. $\endgroup$ – David G. Stork Jul 26 '19 at 19:01
  • 1
    $\begingroup$ 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. $\endgroup$ – DonAntonio Jul 26 '19 at 20:25
  • 2
    $\begingroup$ @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. $\endgroup$ – imranfat Jul 28 '19 at 16:13
  • 2
    $\begingroup$ where are you getting lost with the partial fractions? $\endgroup$ – Doug M Sep 26 '19 at 1:20
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ How exactly does Heaviside method work and where can I read up on it? $\endgroup$ – user671231 Jul 26 '19 at 19:26
  • $\begingroup$ @StackUpPhysics See here. $\endgroup$ – J.G. Jul 26 '19 at 21:10
  • 1
    $\begingroup$ @J.G. Thanks for the link $\endgroup$ – user671231 Jul 26 '19 at 21:14
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.