# Math Help - Integration by Partial Fraction Decomposition

1. ## Integration by Partial Fraction Decomposition

I can find the function that needs to be integrated and get the same function as in the answers but I can’t seem to find a way to integrate the function. I have tried many times and failed, so could someone help me through or show me how its done as I am out of ideas.

2. $\frac{1}{x^3-1}=\frac{1}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$

Now you have to find A, B, C.

3. Hi red_dog

I have said I can already find the function that needs to be integrated but I can’t seem to find a way to integrate the function. That is I have found the values of A, B and C. I’m just not sure how to integrate it when I have this function. Look at the attachment please to see what I mean.

4. $\frac{1}{3}\int\frac{1}{x-1}dx-\frac{1}{3}\int\frac{x+2}{x^2+x+1}dx=$

$=\frac{1}{3}\ln|x-1|-\frac{1}{6}\int\frac{2x+4}{x^2+x+1}dx=$

$=\frac{1}{3}\ln|x-1|-\frac{1}{6}\left(\int\frac{2x+1}{x^2+x+1}dx+3\int\ frac{dx}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} \right)=$

$=\frac{1}{3}\ln|x-1|-\frac{1}{6}\ln(x^2+x+1)-\frac{1}{2}\cdot\frac{2}{\sqrt{3}}\arctan\left(\fr ac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C=$

$=\frac{1}{3}\ln|x-1|-\frac{1}{6}\ln(x^2+x+1)-\frac{1}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}+C$

5. Thanks red_dog

I tried doing something similar to by completing the square but this was where I ran into the trouble. I should be able to the other questions that are similar to this so thanks for the help.