Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
$x-2 = (x+1) - 3$. The first gives a logarithm, the second an $\arctan$. –  Daniel Fischer Dec 10 '13 at 0:52
    
Ahhhh how did I miss that?! Thank you! :D –  yublu Dec 10 '13 at 0:53

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
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. –  yublu Dec 10 '13 at 1:03
    
Looks good to me! –  Igor Rivin Dec 10 '13 at 1:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.