Another partial fraction

**symstar** · September 14th 2008, 06:17 PM

$\int\frac{1}{x^3-1}dx$

Fraction work:
$\frac{1}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$
$1=(A+B)x^2+(A-B+C)x+(A-C)$
$A=\tfrac{1}{3}; B=\tfrac{-1}{3}; C=\tfrac{2}{3}$

Back to the integral:
$\frac{1}{3}\int\frac{dx}{x-1}+\int\frac{\tfrac{-1}{3}x+\tfrac{2}{3}}{x^2+x+1}dx$
$\tfrac{1}{3}\ln{\left|x-1\right|}-\frac{1}{3}\int\frac{xdx}{x^2+x+1}+\frac{2}{3}\int \frac{dx}{x^2+x+1}$

I considered $u=x^2+x+1 \text{ ,but that gives } du=2x+1dx$ which doesn't seem to help the situation much. Should completing the square be a consideration?

**Jhevon** · September 14th 2008, 06:21 PM

Originally Posted by symstar

$\int\frac{1}{x^3-1}dx$

Fraction work:
$\frac{1}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$
$1=(A+B)x^2+(A-B+C)x+(A-C)$
$A=\tfrac{1}{3}; B=\tfrac{-1}{3}; C=\tfrac{2}{3}$

Back to the integral:
$\frac{1}{3}\int\frac{dx}{x-1}+\int\frac{\tfrac{-1}{3}x+\tfrac{2}{3}}{x^2+x+1}dx$
$\tfrac{1}{3}\ln{\left|x-1\right|}-\frac{1}{3}\int\frac{xdx}{x^2+x+1}+\frac{2}{3}\int \frac{dx}{x^2+x+1}$

I considered $u=x^2+x+1 \text{ ,but that gives } du=2x+1dx$ which doesn't seem to help the situation much. Should completing the square be a consideration?

yes, complete the square for the last two integrals

**symstar** · September 14th 2008, 06:37 PM

Take just the integral: $-\frac{1}{3}\int\frac{xdx}{x^2+x+1}$

Completing the square:

$-\frac{1}{3}\int\frac{xdx}{(x+\tfrac{1}{2})^2+\tfra c{3}{4}}$
$u=x+\tfrac{1}{2} \text{ so } du=dx \text{ and }x=u-\tfrac{1}{2}$
$-\frac{1}{3}\int\frac{u-\tfrac{1}{2}du}{u^2+\tfrac{3}{4}}$

Is this correct? This seems to just be leading me in to more of a mess.

**Jhevon** · September 14th 2008, 06:46 PM

Originally Posted by symstar

Take just the integral: $-\frac{1}{3}\int\frac{xdx}{x^2+x+1}$

Completing the square:

$-\frac{1}{3}\int\frac{xdx}{(x+\tfrac{1}{2})^2+\tfra c{3}{4}}$
$u=x+\tfrac{1}{2} \text{ so } du=dx \text{ and }x=u-\tfrac{1}{2}$
$-\frac{1}{3}\int\frac{u-\tfrac{1}{2}du}{u^2+\tfrac{3}{4}}$

Is this correct? This seems to just be leading me in to more of a mess.

yes, you are good so far. now factor out the 3/4 from the denominator and split the integral.

**javax** · September 14th 2008, 06:58 PM

Or with $-\frac{1}{3}\int{\frac{x dx}{x^2+x+1 }}$ you could have go
$<br /> -\frac{1}{3\cdot \color{red}{2}}\int{\frac{\color{red}{2}\color{bla ck}{x+1}}{x^2+x+1}}dx-\int{\frac{dx}{x^2+x+1}}<br />$
I believe now is easier

**symstar** · September 14th 2008, 07:21 PM

Carrying on...
$-\frac{1}{3}\left(\int\frac{u}{u^2+\tfrac{3}{4}}du-\frac{1}{2}\int\frac{du}{u^2+\tfrac{3}{4}}\right)$
$-\frac{1}{3}\left(\frac{1}{2}\ln{(u^2+\tfrac{3}{4}) }-\frac{1}{2}(\frac{2}{\sqrt{3}})\tan^{-1}(\tfrac{2}{\sqrt{3}}(u^2+\tfrac{3}{4})\right)$
$-\tfrac{1}{6}\ln{(x^2+x+1)}-\tfrac{1}{\sqrt{3}}\tan^{-1}(\tfrac{2}{\sqrt{3}}(x^2+x+1)$

If I did indeed do this correctly, then I still have the $\frac{2}{3}\int\frac{dx}{x^2+x+1}$ from earlier which I still need to integrate. According to the book, I'll get the incorrect answer

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