# Partial Fraction Decomposition with an irreducible denominator

How can you use PFD (Partial Fraction Decomposition) to solve this integration problem? I can normally do these fine with denominators that factor, but this one doesn't seem to be able to do that. $$\int_0^2\frac{2x^2+x+8}{x^4+8x+16} \mathrm{d}x$$ Could I set it up as $\displaystyle\frac{Ax^3+Bx^2+Cx+D}{x^4+8x+16}$ and then find the values of $A$,$B$,$C$, and $D$ to complete the integration?

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Perhaps it's a typo and should be $x^4 + 8x^2 + 16 = (x^2+4)^2$. – Bill Dubuque Feb 28 '12 at 20:02
You could set it up as described but the only thing you will find is A=0,B=2,C=1,D=8 so you're back to square one.I belive @BillDubuque is right and it's a typo. – chemeng Feb 28 '12 at 20:05
Yes, you could; and the answer you'll get is $A=0$, $B=2$, $C=1$, and $D=8$; that is, you will just rewrite what you did. However: over the real numbers, any polynomial of degree greater than 2 must factor. Your polynomial does factor. It factors as a product of two irreducible quadratics. – Arturo Magidin Feb 28 '12 at 20:12

Hint: assuming the denominator should be $(x^2+4)^2,\:$ rewrite the numerator as $\rm\:2\:(x^2+4) + x$. Then each summand/denom integrates easily, viz. an arctan + rational function.