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?
Partial Fraction Decomposition with an irreducible denominator
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integration
partial-fractions
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1Perhaps it's a typo and should be $x^4 + 8x^2 + 16 = (x^2+4)^2$. – 2012-02-28
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0You 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. – 2012-02-28
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1Yes, 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. – 2012-02-28
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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.
Almost surely this was what was intended since the numerator has form that enables said easy partial fraction decomposition when the denominator is as speculated.