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I came across this $$\int \frac{dx}{x(x^2+1)^2}$$

in "Method of partial functions" in my Calculus I book. The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way: $$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$

So is there a quicker or a more practical way that I can use?

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  • $\begingroup$ do you mean this here? $$\frac{dx}{x(x^2+1)^2} $$ $\endgroup$ Jan 13, 2017 at 17:11
  • $\begingroup$ What is wrong with the partial fraction? You can get the coefficients quite easy by Residue-Method en.wikipedia.org/wiki/… $\endgroup$
    – Laray
    Jan 13, 2017 at 17:38
  • $\begingroup$ What would that bring us to? $\endgroup$
    – Huzo
    Jan 13, 2017 at 17:50
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    $\begingroup$ $\frac{x^2-x^2+1}{(x^2+1)^2x}$=$-\frac{x}{(x^2+1)^2}+\frac{1}{(x^2+1)x}$ do you see how to proceed? $\endgroup$
    – tired
    Jan 13, 2017 at 17:55
  • $\begingroup$ I love it! Thanks $\endgroup$
    – Huzo
    Jan 13, 2017 at 17:58

5 Answers 5

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Let's take the decomposition you give and see how quick we can go

$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$

Multiplying by $x$ the two sides and making $x=0$ one gets $A=1$.

Multiplying by $(x^2+1)^2$ and making $x=i$ we get $-i=Di+E$ and this means $D=-1$ and $E=0$ . Now rewrite the decomposition

$$\frac{1}{x(x^2+1)^2}= \frac{1}{x}+\frac{Bx+C}{x^2+1}-\frac{x}{(x^2+1)^2}$$

This is equivalent to

$$\begin{align}{Bx+C\over x^2+1}=&{1\over x(x^2+1)^2}-{1\over x}+{x\over (x^2+1)^2}\\=&{(x^2+1)(1-1-x^2)\over x(x^2+1)^2}\\=&-{x\over x^2+1}\end{align}$$

Looks pretty quick to me

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  • $\begingroup$ Wow! My teacher never showed us this way. Very practical indeed $\endgroup$
    – Huzo
    Jan 13, 2017 at 17:56
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    $\begingroup$ Could you explain how you know it's acceptable to just plug in any number and compute results? In particular, $x = 0$ is forbidden by the original expression and we probably want to solve this for the reals, not complex. How do you know going outside these constraints doesn't give an incorrect answer? (I'm not doubting your correctness, but it seems there's nontrivial logic you're not showing.) $\endgroup$
    – jpmc26
    Jan 13, 2017 at 22:00
  • $\begingroup$ Before plugging you multiply by the polynomial that takes away the pole (i.e. The zero from the denominator) for instance multiplying by $x$ the identity become $A+x((Bx+C)/(x^2+1)+(Dx+E)/(x^2+1)^2)=1/(x^2+1)^2$ and this is true for all $x\in \Bbb{C}$, so make $x=0$ you get $A=1$ $\endgroup$
    – marwalix
    Jan 14, 2017 at 8:47
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One way to simplify this a bit, is to multiply numerator and denominator of your integrand by $x$: $$\int \frac{x}{x^2(x^2+1)^2}dx$$

Now substituting $y = x^2$; $dy = 2x dx$, you get the integral $$\frac{1}{2}\int \frac{1}{y(y+1)^2}dy$$ Splitting this in partial fractions is more straightforward than what you had before: $$\frac{1}{y(y+1)^2} = -\frac{1}{(y+1)^2} -\frac{1}{y+1} + \frac{1}{y}$$ Solving the integral then gives $$\frac{1}{2}\left(\frac{1}{y+1} + \log(y) -\log(y+1)\right)+c$$ Subsituting $y = x^2$ then gives $$\frac{1}{2}\left(\frac{1}{x^2+1} + \log(x^2) -\log(x^2+1)\right)+c$$ which can be slightly simplified to $$\frac{1}{2}\left(\frac{1}{x^2+1} -\log(x^2+1)\right)+ \log(x)+c$$

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  • $\begingroup$ This is something I was looking for, thanks $\endgroup$
    – Huzo
    Jan 13, 2017 at 17:45
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Let $\displaystyle \mathcal{I} = \int\frac{1}{x(x^2+1)^2}dx$

substitute $\displaystyle x = \frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}$

So $\displaystyle \mathcal{I} = -\int\frac{t^5}{(1+t^2)^2}\cdot \frac{1}{t^2}dt = -\frac{1}{2}\int\frac{t^2\cdot 2t }{(1+t^2)^2}dt$

put $1+t^2=u\;,$ then $2tdt = du$

So $\displaystyle \mathcal{I} = -\int\frac{(u-1)}{u^2}du = +\int u^{-2}du+\int \frac{1}{u}du$

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  • $\begingroup$ Thanks. A different perspective to the issue which works well. $\endgroup$
    – Huzo
    Jan 13, 2017 at 17:52
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\begin{equation} \int \frac{dx}{x(x^2+1)^2} \end{equation}

\begin{equation} \frac{1}{x(x+i)^2(x-i)^2}=\dfrac{A}{x}+\dfrac{B}{x+i}+\dfrac{C}{x-i}+\dfrac{D}{(x+i)^2}+\dfrac{E}{(x-i)^2} \end{equation}

  1. Let $x=0$. Then $A=\dfrac{1}{(0+i)^2(0-i)^2}=1$\
  2. Let $x=-i$. Then $B=\dfrac{1}{-i(-i-i)^2}=-\dfrac{1}{2}$\
  3. Let $x=i$. Then $C=\dfrac{1}{i(i+i)^2}=-\dfrac{1}{2}$\

Therefore we have

\begin{eqnarray} \frac{1}{x(x^2+1)^2}&=&\dfrac{1}{x}-\dfrac{1}{2}\left(\dfrac{1}{x+i}+\dfrac{1}{x-i}\right)+\dfrac{D}{(x+i)^2}+\dfrac{E}{(x-i)^2}\\ &=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{D(x^2-2ix-1)+E(x^2+2ix-1)}{(x^2+1)^2}\\ &=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{(D+E)x^2+2(E-D)ix-(D+E)}{(x^2+1)^2}\\ &=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{2Eix}{(x^2+1)^2} \end{eqnarray} Note that since the $x^2$ term must vanish it must be the case that $D=-E$.

Thus we only have to find the value of $E$.

\begin{eqnarray} \frac{1}{x(x^2+1)^2}&=&\dfrac{1}{x}-\dfrac{x}{x^2+1}+\dfrac{2Eix}{(x^2+1)^2} \end{eqnarray}

This equation holds for every value of $x$ with the exception of $x=0,i,-i$. Therefore, when $x=1$ it is true that

\begin{eqnarray} \frac{1}{4}&=&1-\dfrac{1}{2}+\dfrac{E}{2}i\\ E&=&\dfrac{i}{2} \end{eqnarray}

Therefore \begin{equation} \frac{1}{x(x^2+1)^2}=\dfrac{1}{x}-\dfrac{x}{x^2+1}-\dfrac{x}{(x^2+1)^2} \end{equation}

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  • $\begingroup$ Does not seem impractical. Thanks! $\endgroup$
    – Huzo
    Jan 13, 2017 at 18:50
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This is not a complete answer, but rather a helpful observation.

First note that the integrand is odd.

Secondly, note that for real factorisations, there is no conjugation from negative terms that can exist (e.g. $1-x$ and $1+x$)

With this in mind, we proceed as follows:

$$\frac{1}{x(x^2+1)^2} = \frac{a}{x} + \frac{bx}{x^2+1}+\frac{cx}{(x^2+1)^2}$$

This reduces the $5$ variable problem into a $3$ variable problem, which is much more manageable.

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