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Here is a partial fractions question from page 287, q22 of Schaum's Calculus 5e:

$$ \int{ \frac{x^6 + 7\, x^5 + 15\, x^4 + 23\, x^2 + 25\, x - 3}{{\left(x^2 + 1\right)}^2\, {\left(x^2 + x + 2\right)}^2} dx} $$

Their answer: (verified)

$$\ln\!\left(\frac{x^2 + 1}{x^2 + x + 2}\right) + \frac{1}{\left(x^2 + x + 2\right)} - \frac{3}{\left(x^2 + 1\right)}$$

The partial fractions decomposition: (verified)

$$ \frac{\left(6\, x + 16\right)}{{\left(x^2 + 1\right)}^2} + \frac{\left(26\, x - 32\right)}{\left(x^2 + 1\right)} - \frac{\left(26\, x - 7\right)}{\left(x^2 + x + 2\right)} + \frac{\left(6\, x + 47\right)}{{\left(x^2 + x + 2\right)}^2} $$

(When I say "verified" I mean I used a math algebra package to verify it on a computer)

Even when working from the computer-done partial fractions decomposition, I still don't get the same answer. What gives? Why is their answer so simple (and correct?) How did they get it? Where's the trick??

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  • $\begingroup$ Please use displayed equations for complex formulas with numerators and denominators; it makes them larger. You do this by placing them within double dollar signs instead of single dollar signs. $\endgroup$
    – Arturo Magidin
    Apr 23, 2011 at 18:38
  • $\begingroup$ Care to tell us what answer you do get? $\endgroup$
    – Arturo Magidin
    Apr 23, 2011 at 18:52
  • $\begingroup$ LoL, looks like a messy one. $\endgroup$
    – night owl
    Apr 23, 2011 at 19:04

1 Answer 1

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The term $\rm\:32\ x^3\:$ is mistakenly omitted in the numerator of the integrand. Once you add that in then the partial fraction is

$$\rm - \frac{2\:x+1}{(x^2+x+2)^2} - \frac{2\:x+1}{x^2+x+2} + \frac{6\:x}{(x^2+1)^2} + \frac{2\:x}{x^2+1} $$

from which the claimed result follows easily.

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