# Question 5b448

Mar 15, 2017

$\frac{1}{3} \ln | x - 1 | - \frac{1}{6} \ln \left({x}^{2} + x + 1\right) + \frac{1}{\sqrt{3}} a r c \tan \left(\frac{2 x + 1}{\sqrt{3}}\right) + C .$

#### Explanation:

Let, $I = \int \frac{x \mathrm{dx}}{{x}^{3} - 1} = \int \frac{x}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} \mathrm{dx} .$

Let us decompoose the Integrand using the Partial Fraction as

x/{(x-1)(x^2+x+1)}=A/(x-1)+(Bx+C)/(x^2+x+1); A,B,C in RR.#

We use Heavyside's Method to find, $A , B , C .$

For $A ,$ we cover $\left(x - 1\right)$ from the integrand and plug in $x = 1$ in its

remaining portion, denoted as, ${\left[\frac{x}{{x}^{2} + x + 1}\right]}_{x = 1} .$

$\therefore A = {\left[\frac{x}{{x}^{2} + x + 1}\right]}_{x = 1} = \frac{1}{3.}$

$\therefore \frac{x}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} - \frac{\frac{1}{3}}{x - 1} = \frac{B x + C}{{x}^{2} + x + 1} , \mathmr{and} ,$

$\frac{x - \frac{1}{3} \left({x}^{2} + x + 1\right)}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} = \frac{B x + C}{{x}^{2} + x + 1} .$

$\therefore \frac{- \frac{1}{3} {x}^{2} + \frac{2}{3} x - \frac{1}{3}}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} = \frac{B x + C}{{x}^{2} + x + 1} .$

$\therefore \frac{\left(\cancel{x - 1}\right) \left(- \frac{1}{3} x + \frac{1}{3}\right)}{\left(\cancel{x - 1}\right) \left({x}^{2} + x + 1\right)} = \frac{B x + C}{{x}^{2} + x + 1} .$

$\therefore B = - \frac{1}{3} , C = \frac{1}{3.}$

Therefore, $I = \frac{1}{3} \int \frac{1}{x - 1} \mathrm{dx} + \int \frac{- \frac{1}{3} x + \frac{1}{3}}{{x}^{2} + x + 1} \mathrm{dx}$

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

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

$= \ldots , , \ldots - \frac{1}{6} \int \frac{\left(2 x + 1\right) - 3}{{x}^{2} + x + 1} \mathrm{dx}$

$= \ldots , , \ldots - \frac{1}{6} \int \frac{2 x + 1}{{x}^{2} + x + 1} \mathrm{dx} + \frac{3}{6} \int \frac{1}{{\left(x + \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} \mathrm{dx}$

$= \ldots , , \ldots - \frac{1}{6} \ln \left({x}^{2} + x + 1\right) + \left(\frac{1}{2}\right) \left(\frac{1}{\frac{\sqrt{3}}{2}}\right) a r c \tan \left\{\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right\} .$

Note that the later integrals were found using the Results :

$\left(1\right) : \int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) + c .$

$\left(2\right) : \int \frac{1}{{\left(x + a\right)}^{2} + {b}^{2}} \mathrm{dx} = \frac{1}{b} a r c \tan \left(\frac{x + a}{b}\right) + c ' \left(b \ne 0\right) .$

Finally, we have,

$I = \frac{1}{3} \ln | x - 1 | - \frac{1}{6} \ln \left({x}^{2} + x + 1\right) + \frac{1}{\sqrt{3}} a r c \tan \left(\frac{2 x + 1}{\sqrt{3}}\right) + C .$

Enjoy Maths.!