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I'm studying for my calc exam(on integration techniques) and i've gone thru the gauntlet of different problems trying to expand my "toolbox". I've run into an integral that I thought I knew how to solve at first, until I looked a little closer. It is as follows:

$$\int\frac{x^5}{x^4-1}\;dx$$

Immediately I long divide the improper integral, yielding(after some LCD work and simplification):

$$\int\frac{x^5}{(x^2+1)(x-1)(x+1)}\;dx$$

To me, this looks like a classic partial fraction decomposition problem, so I begin by setting up my fractions and equating coefficients:

$$\int\frac{x^5}{(x^2+1)(x-1)(x+1)}\;dx\ ;=\; \int(\frac{Ax+B}{x^2+1} + \frac{C}{x-1} + \frac{D}{x+1})\;dx$$

Multiplying by LCD's, I get:

$x^5 \;=\; Ax+B(x^2-1)+ C(x^2+1)(x+1)+ D(x^2+1)(x-1)$ $x^5 \;=\; Ax^3-Ax+Bx^2-B+Cx^3+Cx^2+Cx+C+Dx^3-Dx^2+Dx-D$

I'm pretty confident at this point that my algebra is correct, but I notice a problem; by manipulating the form of the original long division I essentially turned it back into an improper integral, and partial fraction decomposition would yield me more variables than the equation calls for(6 vs the 4 I have A, B, C and D).

So my question is, would would be the best way to tackle this integral? I've learned why I cannot do it this way, but i'm stumped and don't know how to proceed from here.

Original long division yielded: $$\int x+\frac{x}{x^4-1}\;dx$$

I can integrate the $x$ no problem, but i'm not entirely sure how to tackle the $\frac{x}{x^4-1}$. I thought maybe doing a partial fraction decomposition on the $\frac{x}{x^4-1}$ might yield positive results?

Any thoughts or help would be so appreciated. I would like to see my mistakes so I can correct them in the future!

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    $\begingroup$ Do your partial fraction decomposition on $\frac{x}{x^4-1}$ rather than immediately doing partial fraction decomposition on $\frac{x^5}{x^4-1}$. It will work out (you will have four equations in four unknowns and the equations will be consistent). In general when you do partial fraction decomposition the degree of the numerator must be strictly less than the degree of the denominator. $\endgroup$
    – Ian
    Jun 15 '16 at 20:01
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    $\begingroup$ You also have the option of using the substitution $u=x^2$ on $\int\frac{x}{x^4-1}\,dx$. $\endgroup$ Jun 15 '16 at 20:05
  • $\begingroup$ @JohnWaylandBales yes that was exactly what I noticed once I looked over it again. I've been so used to these more complicated techniques that I sometimes forget about good old u-sub :) $\endgroup$ Jun 15 '16 at 20:46
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Setting $u=x^2$, $\mathrm d\mkern1mu u=2x\,\mathrm d\mkern1mu x$, you obtain at once a known integral: $$-\frac12\int\frac{\mathrm d\mkern1mu u}{1-u^2}=\frac14\,\ln\frac{1-u}{1+u}=-\frac12\,\operatorname{artanh} u.$$

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