In integrating $\dfrac{3t^4+2t^2}{(t-1)^3 (t+1)^3}$ I need to decompose this using partial fraction decomposition.

Using Heaviside cover-up method, I am able to easily get the coefficients of $C = \frac58$ and $F = -\frac58$.

$$\frac{3t^4+2t^2}{ (t-1)^3(t+1)^3 } = \frac{A}{t-1} + \frac{B}{(t-1)^2} + \frac{C}{(t-1)^3} + \frac{D}{t+1} + \frac{E}{(t+1)^2} + \frac{F}{(t+1)^3}.$$

I can also get an equation $0 = A + D$ multiplying by $t$ and sending them to infinity.

What would be the fastest and most efficient way of obtaining the rest, $A$, $B$, $D$ and $E$?

Thanks :)

  • $\begingroup$ Ugh. It's quite uncomfortable to read this kind of equations. Please use MathJax. You'll improve the legibility of the equations in less than a minute. $\endgroup$ – Matti P. Mar 26 '20 at 13:35
  • $\begingroup$ Thank you Matti, I am very new to this forum, is there any tutorial on the site? $\endgroup$ – Milan Mar 26 '20 at 13:43

Get a common denominator. Multiply through and equate coefficients of the resulting polynomials.

  • $\begingroup$ Thank you Chris, is there any other way? $\endgroup$ – Milan Mar 26 '20 at 13:47
  • $\begingroup$ That's the only way I know. Sometimes you can cheat a little, and do some of it in your head. Haven't heard of the Heaviside method, but that's what it sounds like. Maybe you could show me. $\endgroup$ – Chris Custer Mar 26 '20 at 13:51
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    $\begingroup$ Typically, I find substituting in nice values for $t$ and solve the resulting linear equations to be much easier than equating coefficients (especially because of calculation errors). $\endgroup$ – Calvin Lin Mar 26 '20 at 14:07
  • $\begingroup$ Of course, you take one denominator root, for example in this case t = 1 and cover (t-1)^3 in the denominator. With this you get coefficient C = 5/8. With this trick you can obtain only the fraction coefficient with the highest power in the denominator, so besides C also F. So I guess its what you referred to with doing in your head. $\endgroup$ – Milan Mar 26 '20 at 14:08

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