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I'm having some trouble solving this integral using partial fraction method: $$\int{\frac{6x}{x^3+8}dx}.$$

After expanding $x^3+8$ into $(x-2)(x^2+2x+4)$ and expanding the original integral into $$\int{\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}}dx,$$ I got $1$,$-1$, and $2$ for $A$, $B$, and $C$, respectively, yielding $$\int{\frac{1}{x-2}+\frac{-x+2}{x^2+2x+4}}dx.$$

This simplifies to $$\ln{|x-2|}-\int{\frac{x-2}{(x+1)^2+3}dx}.$$

How do I solve this resultant integral? That's where I'm stuck.

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    $\begingroup$ $x-2 = (x+1) - 3$. The first gives a logarithm, the second an $\arctan$. $\endgroup$ – Daniel Fischer Dec 10 '13 at 0:52
  • $\begingroup$ Ahhhh how did I miss that?! Thank you! :D $\endgroup$ – akshayc Dec 10 '13 at 0:53
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Write the numerator of the last integral as $(x+1)-3,$ and break the integral up into the two parts. The first integral will be a logarithm the second an arctan but do the calculations yourself.

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  • $\begingroup$ Okay, I got $\ln{|x-2|}-\frac{1}{2}\ln{|(x+1)^2+3|+\sqrt{3}\arctan{(x+1)}}+C$ Is this the right answer? It can be simplified down, obviously, but I want to make sure. $\endgroup$ – akshayc Dec 10 '13 at 1:03
  • $\begingroup$ Looks good to me! $\endgroup$ – Igor Rivin Dec 10 '13 at 1:10

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