Evaluate the following integrals:

$=\int \frac{x}{\left(x^{3}-1\right)} d x=\int \frac{x}{(x-1)\left(x^{2}+x+1\right)} d x$

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

$=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{x-1}{x^{2}+x+1} d x$

$=\frac{1}{3} \log (x-1)-\frac{1}{3}\left[\int \frac{(2 x+1)}{2\left(x^{2}+x+1\right)} d x-\int \frac{3}{2\left(\left(x^{2}+x+1\right)\right)} d x\right]$'

$=\frac{1}{3} \log (x-1)-\frac{1}{3}[I 1+I 2]$

$\mathrm{I}_{1}=\frac{1}{2} \int \frac{(2 x+1)}{\left(x^{2}+x+1\right)} d x$

put $x^{2}+x+1=t$

$(2 x+1) d x=d t$

$\mathrm{I}_{1}=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log t+c=\frac{1}{2} \log \left(x^{2}+x+1\right)+$ c

Now, $I_{2}=\frac{3}{2} \int \frac{d x}{x^{2}+x+1}=\frac{3}{2} \int \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}$

put $(2 x+1) / \sqrt{3}=u$

$2 \mathrm{~d} \times / \sqrt{3}=\mathrm{dt}$

$\mathrm{d} \mathrm{x}=\sqrt{3} \mathrm{dt} / 2$

$=\frac{3}{2} \cdot \frac{2}{\sqrt{3}} \int \frac{d u}{u^{2}+1}=\frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1} u+c=\sqrt{3} \tan ^{-1} \frac{2 x+1}{\sqrt{3}}+c$

So, answer is $=\frac{1}{3} \log (x-1)-\frac{1}{3}\left[\frac{1}{2} \log \left(x^{2}+x+1\right)-\right.$

$\sqrt{3} \tan ^{-1} \frac{2 x+1}{\sqrt{3}}$