itegration of 1/(x^3+1)dx

jazzman

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Jan 20, 2008
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I'm having trouble finding a good method of solving 1x3+1dx\displaystyle \int{\frac{1}{x^3+1}}\,dx.
I tried integration by parts and substitution but haven't succeded in simplifying the expresion...
Please help.
 
Notice the sum of two cubes?.

1x3+1=1(x+1)(x2x+1)\displaystyle \frac{1}{x^{3}+1}=\frac{1}{(x+1)(x^{2}-x+1)}

Now, try partial fractions.

Axx2x+1+Bx2x+1+Cx+1\displaystyle \frac{Ax}{x^{2}-x+1}+\frac{B}{x^{2}-x+1}+\frac{C}{x+1}
 

Thanks a lot! :D
Didn't think of treating it as an integration of a rational function (even though it is such... :p )
 
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