How to i integrate this with out substitutions or Partial fraction decomposition ?
($3x^2$+$2$)/[$x^6$($x^2$+1)]
I've got to : 2/x^6(x^2+1),but after this i haven't been able to eliminate the 2.
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In some sense it seems unproductive to say "Can I integrate this without substitutions or partial fraction decomposition?" While it is important in mathematics to attempt different approaches using alternative ways, it seems to me that these are the very basic methods of evaluating integrals and casting them aside seems to be ignoring the most very basic properties of integrals (and algebra of polynomials, for that matter) – Joshua Mundinger Oct 01 '14 at 13:24
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You can use $\displaystyle\frac{3x^2+2}{x^6(x^2+1)}=\frac{2(x^2+1)}{x^6(x^2+1)}+\frac{x^2}{x^6(x^2+1)}=\frac{2}{x^6}+\frac{1}{x^4(x^2+1)}$
$\displaystyle=\frac{2}{x^6}+\left(\frac{x^2+1}{x^4(x^2+1)}-\frac{x^2}{x^4(x^2+1)}\right)=\frac{2}{x^6}+\frac{1}{x^4}-\frac{1}{x^2(x^2+1)}$
$\displaystyle=\frac{2}{x^6}+\frac{1}{x^4}-\left(\frac{x^2+1}{x^2(x^2+1)}-\frac{x^2}{x^2(x^2+1)}\right)=\frac{2}{x^6}+\frac{1}{x^4}-\frac{1}{x^2}+\frac{1}{x^2+1}.$
[Notice that this gives the partial fraction decomposition of this particular function.]
user84413
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I was struggling to answer with the partial-fraction decomposition, thanks for this. It looks right... – bjd2385 Oct 01 '14 at 20:41
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I like how you split up the numerator, I'll remember that trick in the future... – bjd2385 Oct 01 '14 at 20:52
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HINT:
Use Partial Fraction Decomposition ,
$$\frac{3x^2+2}{x^6(x^2+1)}=\frac{Ax+B}{x^2+1}+\frac Cx+\frac D{x^2}+\cdots+\frac H{x^6}$$
lab bhattacharjee
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I dont want to use partial fraction decomposition, is some other way to integrate it without substitution or PFD ? – questioner Oct 01 '14 at 13:01
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1@user3397, Please mention the prohibited methods clearly in the question itself – lab bhattacharjee Oct 01 '14 at 13:03