Evaluate the following integrals:

Let $I=\int \frac{x}{3 x^{4}-18 x^{2}+11} d x$

Let $\mathrm{x}^{2}=\mathrm{t} \ldots .$ (i)

$\Rightarrow 2 x d x=d t$

$I=\frac{1}{6} \int \frac{1}{t^{2}-6 t+\frac{11}{3}} d t$

$=\frac{1}{6} \int \frac{1}{t^{2}-2 t(3)+(3)^{2}-(3)^{2}+11} d t$

$=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\frac{16}{3}} d t$

Put $t-3=u$.........(ii)

$\Rightarrow \mathrm{dt}=\mathrm{du}$

$I=\frac{1}{6} \int \frac{1}{(u)^{2}-\left(\frac{4}{\sqrt{3}}\right)^{2}} d u$

$I=\frac{1}{6} \times \frac{1}{2 \times \frac{4}{\sqrt{3}}} \log \left|\frac{u-\frac{4}{\sqrt{3}}}{u+\frac{4}{\sqrt{3}}}\right|+c$

$I=\frac{\sqrt{3}}{48} \log \left|\frac{t-3-\frac{4}{\sqrt{3}}}{t-3+\frac{4}{\sqrt{2}}}\right|+c$ [using (ii)]

$I=\frac{\sqrt{3}}{48} \log \left|\frac{x^{2}-3-\frac{4}{\sqrt{3}}}{x^{2}-3+\frac{4}{\sqrt{2}}}\right|+c$ [using (i)]