Solve the integral 1/(1+x^3) dx

  • Thread starter Alexx1
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  • #1
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I try to solve the integral 1/(1+x^3) dx , but I got stuck

Now I got as solution:

(1/3) ln(x+1) - (1/6) ln(x^2 -x+1) + (1/2) integral (1/(x^2 -x+1))

Can someone help me to solve the last integral? I have absolutely no idea how I can solve that one..

So integral: 1/(x^2 -x+1) dx
 

Answers and Replies

  • #2
arildno
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Write:
[tex]\frac{1}{x^{2}-x+1}=\frac{1}{(x-\frac{1}{2})^{2}+\frac{3}{4}}=\frac{4}{3}\frac{1}{(\frac{2x-1}{\sqrt{3}})^{2}+1}[/tex]
 
  • #3
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Write:
[tex]\frac{1}{x^{2}-x+1}=\frac{1}{(x-\frac{1}{2})^{2}+\frac{3}{4}}=\frac{4}{3}\frac{1}{(\frac{2x-1}{\sqrt{3}})^{2}+1}[/tex]

Thank you very much!
 

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