Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$

Question

This is a repost of a question i had written incorrectly earlier.

How do I integrate this without substitutions ?

$$ \frac{x^2+3}{x^6\left(x^2+1\right)} $$

I got: $$ \frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)}, $$ but wasn't able to eliminate the 2.

Related : http://math.stackexchange.com/questions/953936/integration-without-substitution — lab bhattacharjee, Oct 01 '14 at 16:38

score 7 · Answer 1 · answered Oct 01 '14 at 15:26

7

$$\frac{x^2+3}{x^6(x^2+1)}=\frac{2x^4-2x^2+3}{x^6}-\frac{2}{x^2+1}=2x^{-2}-2x^{-4}+3x^{-6}-2\cdot\frac{1}{x^2+1}.$$

answered Oct 01 '14 at 15:26

mathlove

139,939

1

This is good, but the $\frac{1}{x^2+1}$ does require a trig substitution, so it's technically cheating! – beep-boop Oct 01 '14 at 15:29
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...unless you already know that $\int \frac{1}{x^2+1} \mathrm{d}x=\arctan(x)+C$. – beep-boop Oct 01 '14 at 15:29
@mathlove how did you go from step 1 to 2 ? – questioner Oct 01 '14 at 15:35
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@questioner: I just set $\frac{x^2+3}{x^6(x^2+1)}=\frac{ax^4+bx^3+cx^2+dx+e}{x^6}+\frac{f}{x^2+1}$ and compare them. – mathlove Oct 01 '14 at 15:37
@alexqwx Provided you know $$\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$$no need substitution – Anastasiya-Romanova 秀 Oct 01 '14 at 15:38
@mathlove compare how? im sorry if this sounds dumb, i know about this concept – questioner Oct 01 '14 at 15:43
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@questioner: No worry. You can multiply the both sides of the equation I wrote in the above comment by $x^6(x^2+1)$ and you'll get $x^2+3=(ax^4+bx^3+cx^2+dx+e)(x^2+1)+fx^6=(a+f)x^6+bx^5+(a+c)x^4+bx^3+(c+e)x^2+dx+e$. So, you can solve $a+f=0, b=0,a+c=0,b=0,c+e=1,d=0,e=3$ to get $(a,b,c,d,e,f)=(2,0,-2,0,3,-2)$. – mathlove Oct 01 '14 at 15:46
@mathlove so you have used partial fraction decomposition ? – questioner Oct 01 '14 at 16:07
@questioner: Yes, exactly. – mathlove Oct 01 '14 at 16:08
@mathlove Do you think this question could be solved purely through algebraic manipulation (also without substitution) ? – questioner Oct 01 '14 at 16:13
@questioner: Well, you can solve it very easily by what I wrote, can't you? (Or do you want another way?) – mathlove Oct 01 '14 at 16:16
I can, but this question is from the section in my book that suggests integrating the given sums using only algebraic manipulations – questioner Oct 01 '14 at 16:21
@questioner: I think 'partial fraction decomposition' is one of the 'algebraic manipulations'. Don't you think so? – mathlove Oct 01 '14 at 16:24
@mathlove haha it is although i've not yet reached integrating using Partial fraction decomposition in the book so i dont think they'd ask questions using that it. (the question set is before that topic) – questioner Oct 01 '14 at 16:30
@questioner: I see. I hope you learn a bit about partial fraction decomposition from my answer. By the way, Barry Cipra gives a nice answer, which you may like:) – mathlove Oct 01 '14 at 16:33
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I definitely did, thanks! – questioner Oct 01 '14 at 16:36
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@questioner: You are welcome! – mathlove Oct 01 '14 at 16:52

score 7 · Accepted Answer · answered Oct 01 '14 at 16:22

If you don't want to go the formal partial fractions route, you can systematically chip away at the denominator as follows:

$$\begin{align} {x^2+3\over x^6(x^2+1)}&={3(x^2+1)-2x^2\over x^6(x^2+1)}\\ &={3\over x^6}-{2\over x^4(x^2+1)}\\ &={3\over x^6}-{2(x^2+1)-2x^2\over x^4(x^2+1)}\\ &={3\over x^6}-{2\over x^4}+{2\over x^2(x^2+1)}\\ &={3\over x^6}-{2\over x^4}+{2(x^2+1)-2x^2\over x^2(x^2+1)}\\ &={3\over x^6}-{2\over x^4}+{2\over x^2}-{2\over x^2+1}\\ \end{align}$$

The final batch of terms are all easy to integrate, provided you recognize $1/(x^2+1)$ as the derivative of $\arctan x$.

Note: A solid understanding of partial fractions lets you write

$${x^2+3\over x^6(x^2+1)}={Ax^4+Bx^2+C\over x^6}+{D\over x^2+1}$$

(because $x$ appears only to even powers in the expression you're trying to decompose), and then you can solve for the coefficients $A$, $B$, $C$, and $D$. You get the same answer.

score 2 · Answer 3 · answered Oct 01 '14 at 16:56

Once you have the expression equal to $\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)}$ (as you stated in the question), you can get to the partial fraction decomposition without partial fractions. Using the formula for an infinite geometric series, expand $\frac{1}{x^2+1}=1-x^2+x^4-x^6\dots$, which allows you to write your expression as $$\frac{1}{x^6}+\frac{2}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-2(1-x^2+x^4-x^6\dots)=\frac{3}{x^6}-\frac{2}{x^4}+\frac{2}{x^2}-\frac{2}{1+x^2}.$$

Nice! Never thought about the 1/(x^2+1) being an infinite geometric series. — questioner, Oct 01 '14 at 17:06

juantheron · Answer 4 · 2015-09-15T08:40:27.283

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$\bf{My\; Solution}$ Given $\displaystyle \int\frac{x^2+3}{x^6(x^2+1)}dx = \int\frac{(x^2+1)+2}{x^6(x^2+1)}dx = \int x^{-6}dx+\int\frac{2}{x^6(x^2+1)}dx$

$\displaystyle =-\frac{1}{5}x^{-5}+I,$ where $\displaystyle I = \int\frac{2}{x^6(x^2+1)}dx$

Now Using $\displaystyle x= \frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}$. So Let $\displaystyle I = -\int \frac{2t^6}{1+t^2}dt = -\int \frac{2(t^6+1)-2}{1+t^2}dt$

So $\displaystyle I = -2\int \frac{t^6+1}{t^2+1}dt+\int \frac{2}{1+t^2}dt = -2\int (t^4-t^2+1)dt+2\tan^{-1}(t)$

$\displaystyle = -\frac{2t^5}{5}+\frac{2t^3}{3}-\frac{2t^2}{2}+2\tan^{-1}(t)+\mathcal{C} = -\frac{2}{5x^5}+\frac{2}{3x^3}-\frac{2}{2x^2}+2\tan^{-1}\left(\frac{1}{x}\right)+\mathcal{C}$

So $\displaystyle \displaystyle \int\frac{x^2+3}{x^6(x^2+1)}dx = -\frac{1}{5x^5}-\frac{2}{5x^5}+\frac{2}{3x^3}-\frac{2}{2x^2}+2\tan^{-1}\left(\frac{1}{x}\right)+\mathcal{C}$

edited Sep 15 '15 at 08:40

answered Oct 01 '14 at 16:22

juantheron

53,015

Oh dear, I fixed one error that caught my eye, but I overlooked two others: $x^2+3\not=(x^2+1)+1$ at the very beginning, and $\int 1dt\not=t^2/2$ toward the end. I'll delete this comment if juantheron (or someone) fixes things. – Barry Cipra Oct 01 '14 at 16:43
I only just now saw the edit. It looks like you corrected the first error I mentioned but not the second. Note that $t^4-t^2+1$ is an even function of $t$, so it should integrate to an odd function. – Barry Cipra Nov 21 '15 at 14:33

Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$

4 Answers4