Question

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

Answer 1

$$\text{We have},$$
$$I=\int \frac{\text{ x dx}}{x^3-1}$$
$$=\int \frac{\text{ x dx}}{(x-1)(x^2+x+1)}$$
$$\text{ Let}\frac{x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$$
$$\Rightarrow \frac{x}{(x-1)(x^2+x+1)}=\frac{A(x^2+x+1)+(Bx+C)(x-1)}{(x-1)(x^2+x+1)}$$
$$\Rightarrow x=A(x^2+x+1)+B{x}^2-Bx+Cx-C$$
$$\Rightarrow x=(A+B)x^2+(A-B+C)x+A-C$$
$$\text{Equating Coefficient of like terms}$$
$$A+B=0.....\left(1\right)$$
$$A-B+C=1.....\left(2\right)$$
$$A-C=0.....\left(3\right)$$
$$\text{Solving}\left(1\right),\left(2\right)\text{ and }\left(3\right),\text{we get}$$
$$A=\frac{1}{3}$$
$$B=-\frac{1}{3}$$
$$C=\frac{1}{3}$$
$$\therefore \frac{x}{(x-1)(x^2+x+1)}=\frac{1}{3(x-1)}+\frac{-\frac{1}{3}x+\frac{1}{3}}{x^2+x+1}$$
$$=\frac{1}{3(x-1)}+\frac{1}{3}\left[\frac{-x+1}{x^2+x+1}\right]$$
$$=\frac{1}{3(x-1)}-\frac{1}{3}\left[\frac{x-1}{x^2+x+1}\right]$$
$$=\frac{1}{3(x-1)}-\frac{1}{6}\left[\frac{2x-2}{x^2+x+1}\right]$$
$$=\frac{1}{3(x-1)}-\frac{1}{6}\left[\frac{2x+1}{x^2+x+1}\right]-\frac{1}{6}\times \frac{-3}{x^2+x+1}$$
$$=\frac{1}{3(x-1)}-\frac{1}{6}\left[\frac{2x+1}{x^2+x+1}\right]+\frac{1}{2}\times \frac{1}{x^2+x+1}$$
$$\therefore I=\frac{1}{3}\int \frac{dx}{x-1}-\frac{1}{6}\int \frac{(2x+1)dx}{x^2+x+1}+\frac{1}{2}\int \frac{dx}{x^2+x+\frac{1}{4}-\frac{1}{4}+1}$$
$${\text{ Putting x}}^2+x+1=t$$
$$\Rightarrow (2x+1)dx=dt$$
$$\therefore I=\frac{1}{3}\text{ log }|x-1|-\frac{1}{6}\text{ log}\left|t\right|+\frac{1}{2}\int \frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$
$$=\frac{1}{3}\text{ log}|x-1|-\frac{1}{6}\text{ log}|x^2+x+1|+\frac{1}{2}\left[\frac{2}{\sqrt{3}}{\text{ tan}}^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]+C$$
$$=\frac{1}{3}\text{ log }|x-1|-\frac{1}{6}\text{ log}|x^2+x+1|+\frac{1}{\sqrt{3}}{\text{ tan}}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$$